Selection of roots belonging to a segment trigonometry. Learning to select roots of a trigonometric equation

At your request!

13. Solve the equation 3-4cos 2 x=0. Find the sum of its roots belonging to the interval .

Let's reduce the degree of cosine using the formula: 1+cos2α=2cos 2 α. We get an equivalent equation:

3-2(1+cos2x)=0 ⇒ 3-2-2cos2x=0 ⇒ -2cos2x=-1. We divide both sides of the equality by (-2) and get the simplest trigonometric equation:

14. Find b 5 of the geometric progression if b 4 =25 and b 6 =16.

Each term of the geometric progression, starting from the second, is equal to the arithmetic mean of its neighboring terms:

(b n) 2 =b n-1 ∙b n+1 . We have (b 5) 2 =b 4 ∙b 6 ⇒ (b 5) 2 =25·16 ⇒ b 5 =±5·4 ⇒ b 5 =±20.

15. Find the derivative of the function: f(x)=tgx-ctgx.

16. Find the greatest and smallest value functions y(x)=x 2 -12x+27

on the segment.

To find the largest and smallest values ​​of a function y=f(x) on the segment, you need to find the values ​​of this function at the ends of the segment and at those critical points that belong to this segment, and then select the largest and smallest from all the obtained values.

Let's find the values ​​of the function at x=3 and at x=7, i.e. at the ends of the segment.

y(3)=3 2 -12∙3+27 =9-36+27=0;

y(7)=7 2 -12∙7+27 =49-84+27=-84+76=-8.

Find the derivative of this function: y’(x)=(x 2 -12x+27)’ =2x-12=2(x-6); the critical point x=6 belongs to this interval. Let's find the value of the function at x=6.

y(6)=6 2 -12∙6+27 =36-72+27=-72+63=-9. Now we choose from the three obtained values: 0; -8 and -9 largest and smallest: at the largest. =0; at name =-9.

17. Find general form antiderivatives for the function:

This interval is the domain of definition of this function. Answers should begin with F(x), and not with f(x) - after all, we are looking for an antiderivative. By definition, the function F(x) is an antiderivative of the function f(x) if the equality holds: F’(x)=f(x). So you can simply find derivatives of the proposed answers until you get it this function. A rigorous solution is the calculation of the integral of a given function. We apply the formulas:

19. Write an equation for the line containing the median BD of triangle ABC if its vertices are A(-6; 2), B(6; 6) C(2; -6).

To compile the equation of a line, you need to know the coordinates of 2 points of this line, but we only know the coordinates of point B. Since the median BD divides the opposite side in half, point D is the midpoint of the segment AC. The coordinates of the middle of a segment are the half-sums of the corresponding coordinates of the ends of the segment. Let's find the coordinates of point D.

20. Calculate:

24. The area of ​​a regular triangle lying at the base of a right prism is equal to

This problem is the inverse of problem No. 24 from option 0021.

25. Find the pattern and insert the missing number: 1; 4; 9; 16; ...

Obviously this number 25 , since we are given a sequence of squares of natural numbers:

1 2 ; 2 2 ; 3 2 ; 4 2 ; 5 2 ; …

Good luck and success to everyone!

The simplest trigonometric equations are solved, as a rule, using formulas. Let me remind you that the simplest trigonometric equations are:

sinx = a

cosx = a

tgx = a

ctgx = a

x is the angle to be found,
a is any number.

And here are the formulas with which you can immediately write down the solutions to these simplest equations.

For sine:

For cosine:

x = ± arccos a + 2π n, n ∈ Z

For tangent:

x = arctan a + π n, n ∈ Z

For cotangent:

x = arcctg a + π n, n ∈ Z

Actually, this is the theoretical part of solving the simplest trigonometric equations. Moreover, everything!) Nothing at all. However, the number of errors on this topic is simply off the charts. Especially if the example deviates slightly from the template. Why?

Yes, because a lot of people write down these letters, without understanding their meaning at all! He writes down with caution, lest something happen...) This needs to be sorted out. Trigonometry for people, or people for trigonometry, after all!?)

Let's figure it out?

One angle will be equal to arccos a, second: -arccos a.

And it will always work out this way. For any A.

If you don’t believe me, hover your mouse over the picture, or touch the picture on your tablet.) I changed the number A to something negative. Anyway, we got one corner arccos a, second: -arccos a.

Therefore, the answer can always be written as two series of roots:

x 1 = arccos a + 2π n, n ∈ Z

x 2 = - arccos a + 2π n, n ∈ Z

Let's combine these two series into one:

x= ± arccos a + 2π n, n ∈ Z

And that's all. We have obtained a general formula for solving the simplest trigonometric equation with cosine.

If you understand that this is not some kind of superscientific wisdom, but just a shortened version of two series of answers, You will also be able to handle tasks “C”. With inequalities, with selecting roots from a given interval... There the answer with a plus/minus does not work. But if you treat the answer in a businesslike manner and break it down into two separate answers, everything will be resolved.) Actually, that’s why we’re looking into it. What, how and where.

In the simplest trigonometric equation

sinx = a

we also get two series of roots. Always. And these two series can also be recorded in one line. Only this line will be trickier:

x = (-1) n arcsin a + π n, n ∈ Z

But the essence remains the same. Mathematicians simply designed a formula to make one instead of two entries for series of roots. That's all!

Let's check the mathematicians? And you never know...)

In the previous lesson, the solution (without any formulas) of a trigonometric equation with sine was discussed in detail:

The answer resulted in two series of roots:

x 1 = π /6 + 2π n, n ∈ Z

x 2 = 5π /6 + 2π n, n ∈ Z

If we solve the same equation using the formula, we get the answer:

x = (-1) n arcsin 0.5 + π n, n ∈ Z

Actually, this is an unfinished answer.) The student must know that arcsin 0.5 = π /6. The complete answer would be:

x = (-1) n π /6+ π n, n ∈ Z

Here it arises interest Ask. Reply via x 1; x 2 (this is the correct answer!) and through lonely X (and this is the correct answer!) - are they the same thing or not? We'll find out now.)

We substitute in the answer with x 1 values n =0; 1; 2; etc., we count, we get a series of roots:

x 1 = π/6; 13π/6; 25π/6 and so on.

With the same substitution in response with x 2 , we get:

x 2 = 5π/6; 17π/6; 29π/6 and so on.

Now let's substitute the values n (0; 1; 2; 3; 4...) into the general formula for single X . That is, we raise minus one to the zero power, then to the first, second, etc. Well, of course, we substitute 0 into the second term; 1; 2 3; 4, etc. And we count. We get the series:

x = π/6; 5π/6; 13π/6; 17π/6; 25π/6 and so on.

That's all you can see.) General formula gives us exactly the same results as are the two answers separately. Just everything at once, in order. The mathematicians were not fooled.)

Formulas for solving trigonometric equations with tangent and cotangent can also be checked. But we won’t.) They are already simple.

I wrote out all this substitution and checking specifically. Here it is important to understand one simple thing: there are formulas for solving elementary trigonometric equations, just a short summary of the answers. For this brevity, we had to insert plus/minus into the cosine solution and (-1) n into the sine solution.

These inserts do not interfere in any way in tasks where you just need to write down the answer to an elementary equation. But if you need to solve an inequality, or then you need to do something with the answer: select roots on an interval, check for ODZ, etc., these insertions can easily unsettle a person.

So what should I do? Yes, either write the answer in two series, or solve the equation/inequality using the trigonometric circle. Then these insertions disappear and life becomes easier.)

We can summarize.

To solve the simplest trigonometric equations, there are ready-made answer formulas. Four pieces. They are good for instantly writing down the solution to an equation. For example, you need to solve the equations:

sinx = 0.3

Easily: x = (-1) n arcsin 0.3 + π n, n ∈ Z

cosx = 0.2

No problem: x = ± arccos 0.2 + 2π n, n ∈ Z

tgx = 1.2

Easily: x = arctan 1,2 + π n, n ∈ Z

ctgx = 3.7

One left: x= arcctg3,7 + π n, n ∈ Z

cos x = 1.8

If you, shining with knowledge, instantly write the answer:

x= ± arccos 1.8 + 2π n, n ∈ Z

then you are already shining, this... that... from a puddle.) Correct answer: there are no solutions. Don't understand why? Read what arc cosine is. In addition, if on the right side of the original equation there are tabular values ​​of sine, cosine, tangent, cotangent, - 1; 0; √3; 1/2; √3/2 and so on. - the answer through the arches will be unfinished. Arches must be converted to radians.

And if you come across inequality, like

then the answer is:

x πn, n ∈ Z

there is rare nonsense, yes...) Here you need to solve using the trigonometric circle. What we will do in the corresponding topic.

For those who heroically read to these lines. I simply cannot help but appreciate your titanic efforts. Bonus for you.)

Bonus:

When writing down formulas in an alarming combat situation, even seasoned nerds often get confused about where πn, And where 2π n. Here's a simple trick for you. In everyone formulas worth πn. Except for the only formula with arc cosine. It stands there 2πn. Two peen. Keyword - two. In this same formula there are two sign at the beginning. Plus and minus. Here and there - two.

So if you wrote two sign before the arc cosine, it’s easier to remember what will happen at the end two peen. And it also happens the other way around. The person will miss the sign ± , gets to the end, writes correctly two Pien, and he’ll come to his senses. There's something ahead two sign! The person will return to the beginning and correct the mistake! Like this.)

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

No. 10 (757) PUBLISHED SINCE 1992 mat.1september.ru Topic of the issue Testing knowledge Our project Competitions Attention - Creative Analysis of the lesson Ural Cup for the strong exam “Axiom of the student of parallel lines” c. 16 p.m. 20 p.m. 44 7 6 5 4 3 version of the magazine ja va l 2 o n a n e r t e l e n t e l n i d o p o t e r a l s 1 m a i n e t b m a c h i n L i t e r u s 1 2 3 4 5 6 0 r. w w be w. 1 m septe October 1september.ru 2014 mathematic Subscription on the website www.1september.ru or through the Russian Post catalog: 79073 (paper version); 12717 (CD version) grades 10–11 Selection training S. MUGALLIMOVA, pos. Bely Yar, Tyumen region. roots of trigonometric equation Trigonometry in school course mathematics occupies a special place and is traditionally considered difficult both for the teacher to present and for the students to master. This is one of the sections, the study of which is often perceived by many as “mathematics for the sake of mathematics,” as the study of material that has no practical value. Meanwhile, the trigonometric apparatus is used in many applications of mathematics, and operating with trigonometric functions is necessary to implement intra- and interdisciplinary connections in teaching mathematics. Note that trigonometric material creates fertile ground for the formation of various meta-subject skills. For example, learning to select roots of a trigonometric equation and solutions trigonometric inequality allows you to develop the skill associated with finding solutions that satisfy the combination of given conditions. The method of teaching root selection is based on the facts listed below. Knowledge of: – location of points on a trigonometric circle; – signs trigonometric functions; – locations of points corresponding to the most common angle values, and angles associated with them by reduction formulas; – graphs of trigonometric functions and their properties. Understanding: – that on a trigonometric circle a point is characterized by three indicators: 1) the angle of rotation of the point P (1; 0); 2) the abscissa, which corresponds to the cosine of this angle and 3) the ordinate, which corresponds to the sine of this angle; – ambiguity of writing the root of a trigonometric equation and the dependence specific meaning root of the value of an integer parameter; – dependence of the radius rotation angle on the number of full revolutions or on the period of the function. Ability to: – mark points on a trigonometric circle corresponding to positive and negative angles of rotation of the radius; – correlate the values ​​of trigonometric functions with the location of a point on a trigonometric circle; mathematics October 2014 – write down the values ​​of the angles of rotation of a point 3.3. Mark as many points as possible that correspond to the symmetrical points corresponding to the given values ​​of the function on the trigonometric circle; 1 (for example | sin x | =). – write down the values ​​of the arguments of trigonometric functions at the points of the graph of the function 3.4. Mark the intervals corresponding to the tion, taking into account the periodicity of the function, as well as the given restrictions on the values ​​of the even and odd function; 3 1 (for example, − ≤ cos x ≤). – using the values ​​of the variables, find the corresponding points on the graphs of functions; 3.5. For given values ​​of the function and limit - combine a series of trigonometric roots on the values ​​of the argument and note the corresponding equations. corresponding points and write down the values ​​of the argument- Thus, in the process of studying the trigonoment (for example, indicate on the graph and make metric material, it is necessary to perform the appropriate notes for the points, the following exercises. 5π satisfying the conditions tg x = 3 and −3π< x <). 1. При изучении начал тригонометрии (в пря- 2 моугольном треугольнике) заполнить (и запом- Перечисленные выше действия полезны при нить!) таблицу значений тригонометрических решении задачи С1 ЕГЭ по математике. В этой функций для углов 30°, 45°, 60° и 90°. задаче, помимо решения тригонометрического 2. При введении понятия тригонометрической уравнения, требуется произвести отбор корней, окружности: и для успешного выполнения этого задания на 2.1. Отметить точки, соответствующие по- экзамене, помимо перечисленных знаний и уме- воротам радиуса на 30°, 45°, 60°, затем на 0, ний, ученик должен владеть следующими навы- π 3π π π π π π π 5π 3π ками: , π, 2π, − , − , − , 2 2 6 4 3 6 4 3 6 4 – решать простейшие тригонометрические 2π 7π 5π 4π уравнения и неравенства; , . 3 6 4 3 – применять тригонометрические тождества; 2.2. Записать значения углов для указанных – использовать различные методы решения выше точек с учетом периодичности движения уравнений; по окружности. – решать двойные линейные неравенства; 2.3. Записать значения углов для указанных – оценивать значение иррационального числа. выше точек с учетом периодичности движения Перечислим способы отбора корней в подоб- по окружности при заданных значениях параме- ных заданиях. тра (например, при n = 2, n = –1, n = –5). 2.4. Найти с помощью тригонометрической Способ перевода в градусную меру окружности значения синуса, косинуса, танген- 1 Найти корни уравнения sin x = , удовлетво- са и котангенса для указанных выше углов. 2 2.5. Отметить на окружности точки, соответ-  3π 5π  ряющие условию x ∈  − ;  . ствующие требуемым значениям тригонометри-  2 2  ческих функций. Решение. Корни уравнения имеют вид 2.6. Записать числовые промежутки, удовлет- π x = (−1)n + πn, где n ∈ Z. воряющие заданным ограничениям значения 6 3 2 Это значит, что функции (например, − ≤ sin α ≤). 2 2 x = 30° + 360°жn или x = 150° + 360°жn. 2.7. Подобрать формулу для записи углов, со-  3π 5π  ответствующих нескольким точкам на тригоно- Условие x ∈  − ; можно записать в виде метрической окружности (например, объединить  2 2  π 3π x ∈ [–270°; 450°]. Указанному промежутку при- записи x = ± + 2πn, n ∈ Z, и x = ± + 2πk, k ∈ Z). 4 4 надлежат следующие значения: 3. При изучении тригонометрических функ- ций, их свойств и графиков: 30°, 150°, –210°, 390°. 3.1. Отметить на графике функции точки, со- Выразим величины этих углов в радианах: ответствующие указанным выше значениям ар- π 5π 7π 13π , − , . гументов. 6 6 6 6 3.2. При заданном значении функции (напри- Это не самый изящный способ решения по- мер, ctg x = 1) отметить как можно больше точек добных заданий, но он полезен на первых порах на графике функции и записать соответствую- освоения действия и в работе со слабыми учени- щие значения аргумента. ками. 31 математика октябрь 2014 Способ движения по окружности Способ оценки 3 Решить уравнение Найти корни уравнения tg x = , удовлетво- tg x − 1 3 = 0.  π  − cos x ряющие условию x ∈  − ; 2π  .  2  Решение. Данное уравнение равносильно си- 3 Решение. Корни уравнения tg x = имеют стеме  tg x = 1, π 3  вид x = + πn, n ∈ Z. Потребуем выполнения 6  cos x < 0.  π  условия x ∈  − ; 2π  , для этого решим двойное Отметим на тригонометрической окружности  2  корни уравнения tg x = 1, соответствующие зна- неравенство: π π π 2 5 чениям углов поворота x = + πn, n ∈ Z (рис. 1). − ≤ + πn ≤ 2π, − ≤ n ≤ 1 . 4 2 6 3 6 Выделим также дуги окружности, лежащие во II π 7π Отсюда n = 0 или n = 1. Значит x = или x = . и III координатных четвертях, так как в этих чет- 6 6 вертях выполнено условие cos x < 0. Графический способ 1 Найти корни уравнения sin x = , удовлетво- 2  3π 5π  ряющие условию x ∈  − ;  .  2 2  Решение. Построим график функции y = sin x (рис. 2). Корни данного уравнения являются абс- циссами точек пересечения графика с прямой практикум 1 y= . Отметим такие точки, выделив фрагмент 2  3π 5π  графика на промежутке  − ;  .  2 2  Рис. 1 Из рисунка видно, что решениями системы, а значит, и решениями данного уравнения явля- / π ются значения x = + π(2n + 1), n ∈ Z. м е то д о б ъ е д и н е н и е 4 Рис. 2 Способ перебора Здесь cos x π π 5π π 13π Решить уравнение = 0. x0 = , x1 = π − = , x2 = + 2π = , 16 − x 2 6 6 6 6 6 Решение. Данное уравнение равносильно си- 5π 7π стеме x3 = − 2π = − . 6 6  cos x = 0,  16 − x >0. 2 Thus, on a given interval the equation has four roots: From the equation cos x = 0 we obtain: x = + πn, n ∈ Z. 2 π 5π 13π 7π , − . The solutions to the inequality 16 – x2 > 0 belong to the 6 6 6 6 interval (–4; 4). In conclusion, let us highlight a few points. Let's perform an exhaustive search: A skill associated with finding solutions that satisfy the given values ​​of the argument, if n = 0, then x = + π ⋅0 = ≈ ∈(−4; 4); 2 2 2 is important in solving many applied problems, and it is necessary to develop this skill - if n = 1, then x = + π = ≈ ∉(−4; 4); 2 2 2 mo in the process of studying everything trigonometrically - if n ≥ 1, then we get values ​​of x greater than 4; Russian material. π π 3, 14 In the process of learning to solve problems, in which - if n = –1, then x = −π= − ≈ − ∈(−4; 4); 2 2 2 it is necessary to select the roots of the trigonometric equation, π 3π 3 ⋅ 3, 14 should be discussed with students if n = –2, then x = − 2π = − ≈− ∉(−4; 4); 2 2 2 different ways performing this action, and if n ≤ –2, then we obtain x values ​​less than –4. also find out the cases when one or another method may be the most convenient or, on- This equation has two roots: and - . 2 2 turn, unusable. mathematics October 2014 32

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This article can help high school students, as well as teachers, in solving trigonometric equations and selecting roots belonging to a certain interval. Depending on what restrictions are given on the roots obtained, you should use different methods for selecting roots, that is, you need to take the method that will more clearly show the correct result.

View document contents
“WAYS OF SELECTING ROOT OF TRIGONOMETRIC EQUATIONS”

METHODS OF SELECTING ROOT OF TRIGONOMETRIC EQUATIONS

Popova Tatyana Sergeevna, teacher of mathematics, computer science, physics MCOU BGO Petrovskaya Secondary School

The Unified State Exam in mathematics includes tasks related to solving equations. There are linear, quadratic, rational, irrational, exponential, logarithmic and trigonometric equations. These equations are required: firstly, to solve, that is, to find all their solutions, and secondly, to select roots belonging to one or another interval. In this article we will consider an example of solving a trigonometric equation and selecting its roots different ways. Depending on what restrictions are given on the roots obtained, you should use different methods for selecting roots, that is, you need to take the method that will more clearly show the correct result.

Let's consider three ways to select roots:

Using a unit circle;

Using inequalities;

Using a graph.

On specific example Let's look at these methods.

Let the following task be given:

a) Solve the equation

b) Indicate the roots of this equation that belong to the segment.

Let's decide first given equation:

Using the double angle formula and the ghost formula, we get:

From here, or. Solving each equation, we get:

; or
.

b) You can select roots using a unit circle (Fig. 1), but children get confused, since the given interval may be greater than the length of the circle and it is difficult to depict it when applied to the circle:

We get the numbers:

You can use the inequality method. Note that if a segment is given, then the inequality is not strict, and if it is an interval, then the inequality is strict. Let's check every root

Considering that -3,-2. Substituting n into the root formula, we get roots ; x=

Similarly, we find the roots for,

k- no whole ones,

1, substitute into the common root

We obtained exactly the same roots as using the unit circle.

This method may be more cumbersome, but own experience While working on solving such equations and selecting roots with students, we noticed that using the inequalities method, schoolchildren make fewer mistakes.

Using the same example, let’s consider selecting the roots of an equation using a graph (Fig. 2)

We also get three roots:

We need to teach children to use all three methods of selecting roots, and then let them decide for themselves which method is easier for them and which method is closer. You can also test yourself whether your decision is correct using different methods.

Used Books:

http://yourtutor.info

http://www.ctege.info/zadaniya-ege-po-matematike