A lesson in the integrated application of knowledge.
Lesson objectives.
- Consider various methods solving trigonometric equations.
- Development creativity students by solving equations.
- Encouraging students to self-control, mutual control, and self-analysis of their educational activities.
Equipment: screen, projector, reference material.
During the classes
Introductory conversation.
The main method for solving trigonometric equations is to reduce them to their simplest form. In this case, they apply usual ways, such as factoring, as well as techniques used only for solving trigonometric equations. There are quite a lot of these techniques, for example, various trigonometric substitutions, angle transformations, transformations of trigonometric functions. The indiscriminate application of any trigonometric transformations usually does not simplify the equation, but catastrophically complicates it. To work out in general outline plan for solving the equation, outline a way to reduce the equation to the simplest, you must first analyze the angles - the arguments of the trigonometric functions included in the equation.
Today we will talk about methods for solving trigonometric equations. The correctly chosen method can often significantly simplify the solution, so all the methods we have studied should always be kept in mind in order to solve trigonometric equations using the most appropriate method.
II. (Using a projector, we repeat the methods for solving equations.)
1. Method of reducing a trigonometric equation to an algebraic one.
It is necessary to express all trigonometric functions through one, with the same argument. This can be done using the basic trigonometric identity and its consequences. We obtain an equation with one trigonometric function. Taking it as a new unknown, we obtain an algebraic equation. We find its roots and return to the old unknown, solving the simplest trigonometric equations.
2. Factorization method.
To change angles, formulas for reduction, sum and difference of arguments are often useful, as well as formulas for converting the sum (difference) of trigonometric functions into a product and vice versa.
sin x + sin 3x = sin 2x + sin4x
3. Method of introducing an additional angle.
4. Method of using universal substitution.
Equations of the form F(sinx, cosx, tanx) = 0 are reduced to algebraic using a universal trigonometric substitution
Expressing sine, cosine and tangent in terms of the tangent of a half angle. This technique can lead to a higher order equation. The solution to which is difficult.
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.) The 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. It is important to understand one thing here simple thing: there are formulas for solving elementary trigonometric equations, only, short note 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.
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When solving many mathematical problems, especially those that occur before grade 10, the order of actions performed that will lead to the goal is clearly defined. Such problems include, for example, linear and quadratic equations, linear and quadratic inequalities, fractional equations and equations that reduce to quadratic ones. The principle of successfully solving each of the mentioned problems is as follows: you need to establish what type of problem you are solving, remember the necessary sequence of actions that will lead to the desired result, i.e. answer and follow these steps.
It is obvious that success or failure in solving a particular problem depends mainly on how correctly the type of equation being solved is determined, how correctly the sequence of all stages of its solution is reproduced. Of course, in this case it is necessary to have the skills to perform identical transformations and calculations.
The situation is different with trigonometric equations. It is not at all difficult to establish the fact that the equation is trigonometric. Difficulties arise when determining the sequence of actions that would lead to the correct answer.
By appearance equation, it is sometimes difficult to determine its type. And without knowing the type of equation, it is almost impossible to choose the right one from several dozen trigonometric formulas.
To solve a trigonometric equation, you need to try:
1. bring all functions included in the equation to “the same angles”;
2. bring the equation to “identical functions”;
3. factor the left side of the equation, etc.
Let's consider basic methods for solving trigonometric equations.
I. Reduction to the simplest trigonometric equations
Solution diagram
Step 1. Express trigonometric function through known components.
Step 2. Find the function argument using the formulas:
cos x = a; x = ±arccos a + 2πn, n ЄZ.
sin x = a; x = (-1) n arcsin a + πn, n Є Z.
tan x = a; x = arctan a + πn, n Є Z.
ctg x = a; x = arcctg a + πn, n Є Z.
Step 3. Find the unknown variable.
Example.
2 cos(3x – π/4) = -√2.
Solution.
1) cos(3x – π/4) = -√2/2.
2) 3x – π/4 = ±(π – π/4) + 2πn, n Є Z;
3x – π/4 = ±3π/4 + 2πn, n Є Z.
3) 3x = ±3π/4 + π/4 + 2πn, n Є Z;
x = ±3π/12 + π/12 + 2πn/3, n Є Z;
x = ±π/4 + π/12 + 2πn/3, n Є Z.
Answer: ±π/4 + π/12 + 2πn/3, n Є Z.
II. Variable replacement
Solution diagram
Step 1. Reduce the equation to algebraic form with respect to one of the trigonometric functions.
Step 2. Denote the resulting function by the variable t (if necessary, introduce restrictions on t).
Step 3. Write down and solve the resulting algebraic equation.
Step 4. Make a reverse replacement.
Step 5. Solve the simplest trigonometric equation.
Example.
2cos 2 (x/2) – 5sin (x/2) – 5 = 0.
Solution.
1) 2(1 – sin 2 (x/2)) – 5sin (x/2) – 5 = 0;
2sin 2 (x/2) + 5sin (x/2) + 3 = 0.
2) Let sin (x/2) = t, where |t| ≤ 1.
3) 2t 2 + 5t + 3 = 0;
t = 1 or e = -3/2, does not satisfy the condition |t| ≤ 1.
4) sin(x/2) = 1.
5) x/2 = π/2 + 2πn, n Є Z;
x = π + 4πn, n Є Z.
Answer: x = π + 4πn, n Є Z.
III. Equation order reduction method
Solution diagram
Step 1. Replace this equation with a linear one, using the formula for reducing the degree:
sin 2 x = 1/2 · (1 – cos 2x);
cos 2 x = 1/2 · (1 + cos 2x);
tg 2 x = (1 – cos 2x) / (1 + cos 2x).
Step 2. Solve the resulting equation using methods I and II.
Example.
cos 2x + cos 2 x = 5/4.
Solution.
1) cos 2x + 1/2 · (1 + cos 2x) = 5/4.
2) cos 2x + 1/2 + 1/2 · cos 2x = 5/4;
3/2 cos 2x = 3/4;
2x = ±π/3 + 2πn, n Є Z;
x = ±π/6 + πn, n Є Z.
Answer: x = ±π/6 + πn, n Є Z.
IV. Homogeneous equations
Solution diagram
Step 1. Reduce this equation to the form
a) a sin x + b cos x = 0 ( homogeneous equation first degree)
or to the view
b) a sin 2 x + b sin x · cos x + c cos 2 x = 0 (homogeneous equation of the second degree).
Step 2. Divide both sides of the equation by
a) cos x ≠ 0;
b) cos 2 x ≠ 0;
and get the equation for tan x:
a) a tan x + b = 0;
b) a tan 2 x + b arctan x + c = 0.
Step 3. Solve the equation using known methods.
Example.
5sin 2 x + 3sin x cos x – 4 = 0.
Solution.
1) 5sin 2 x + 3sin x · cos x – 4(sin 2 x + cos 2 x) = 0;
5sin 2 x + 3sin x · cos x – 4sin² x – 4cos 2 x = 0;
sin 2 x + 3sin x · cos x – 4cos 2 x = 0/cos 2 x ≠ 0.
2) tg 2 x + 3tg x – 4 = 0.
3) Let tg x = t, then
t 2 + 3t – 4 = 0;
t = 1 or t = -4, which means
tg x = 1 or tg x = -4.
From the first equation x = π/4 + πn, n Є Z; from the second equation x = -arctg 4 + πk, k Є Z.
Answer: x = π/4 + πn, n Є Z; x = -arctg 4 + πk, k Є Z.
V. Method of transforming an equation using trigonometric formulas
Solution diagram
Step 1. Using all possible trigonometric formulas, reduce this equation to an equation solved by methods I, II, III, IV.
Step 2. Solve the resulting equation using known methods.
Example.
sin x + sin 2x + sin 3x = 0.
Solution.
1) (sin x + sin 3x) + sin 2x = 0;
2sin 2x cos x + sin 2x = 0.
2) sin 2x (2cos x + 1) = 0;
sin 2x = 0 or 2cos x + 1 = 0;
From the first equation 2x = π/2 + πn, n Є Z; from the second equation cos x = -1/2.
We have x = π/4 + πn/2, n Є Z; from the second equation x = ±(π – π/3) + 2πk, k Є Z.
As a result, x = π/4 + πn/2, n Є Z; x = ±2π/3 + 2πk, k Є Z.
Answer: x = π/4 + πn/2, n Є Z; x = ±2π/3 + 2πk, k Є Z.
The ability and skill to solve trigonometric equations is very 
important, their development requires significant effort, both on the part of the student and on the part of the teacher.
Many problems of stereometry, physics, etc. are associated with the solution of trigonometric equations. The process of solving such problems embodies many of the knowledge and skills that are acquired by studying the elements of trigonometry.
Trigonometric equations occupy an important place in the process of learning mathematics and personal development in general.
Still have questions? Don't know how to solve trigonometric equations?
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It is no secret that success or failure in the process of solving almost any problem mainly depends on the correct definition of the type given equation, as well as from the correct reproduction of the sequence of all stages of its solution. However, in the case of trigonometric equations, determining the fact that the equation is trigonometric is not at all difficult. But in the process of determining the sequence of actions that should lead us to the correct answer, we may encounter certain difficulties. Let's figure out how to solve trigonometric equations correctly from the very beginning.
Solving trigonometric equations
In order to solve a trigonometric equation, you need to try the following points:
- We reduce all the functions that are included in our equation to “identical angles”;
- It is necessary to bring the given equation to “identical functions”;
- We decompose the left side of the given equation into factors or other necessary components.
Methods
Method 1. Such equations must be solved in two stages. First, we transform the equation in order to obtain its simplest (simplified) form. The equation: Cosx = a, Sinx = a and similar ones are called the simplest trigonometric equations. The second stage is solving the simplest equation obtained. It should be noted that the simplest equation can be solved using the algebraic method, which is well known to us from the school algebra course. It is also called the substitution and variable replacement method. Using reduction formulas, you first need to transform, then make a substitution, and then find the roots.
Next, we need to factor our equation into possible factors; to do this, we need to move all terms to the left and then we can factor it. Now we need to bring this equation to a homogeneous one, in which all terms are equal to the same degree, and the cosine and sine have the same angle.
Before solving trigonometric equations, you need to move its terms to the left side, taking them from the right side, and then put all the common denominators out of brackets. We equate our brackets and factors to zero. Our equated brackets represent a homogeneous equation with a reduced degree, which must be divided by sin (cos) to the highest degree. Now we solve the algebraic equation that was obtained in relation to tan.
Method 2. Another method by which you can solve a trigonometric equation is to go to the half angle. For example, we solve the equation: 3sinx-5cosx=7.
We need to go to the half angle, in our case it is: 6sin(x/2)*cos(x/2)- 5cos²(x/2)+5sin²(x/2) = 7sin²(x/2)+7cos²(x /2).And after that, we reduce all the terms into one part (for convenience, it is better to choose the right one) and proceed to solve the equation.
If necessary, you can enter an auxiliary angle. This is done in the case when you need to replace the integer value sin (a) or cos (a) and the sign “a” just acts as an auxiliary angle.
Product to sum
How to solve trigonometric equations using product to sum? A method known as product-to-sum conversion can also be used to solve such equations. In this case, it is necessary to use the formulas corresponding to the equation.
For example, we have the equation: 2sinx * sin3x= сos4x
We need to solve this problem by converting the left side into a sum, namely:
сos 4x –cos8x=cos4x,
x = p/16 + pk/8.
If the above methods are not suitable, and you still do not know how to solve simple trigonometric equations, you can use another method - universal substitution. It can be used to transform an expression and make a substitution. For example: Cos(x/2)=u. Now you can solve the equation with the existing parameter u. And having received the desired result, do not forget to convert this value to the opposite.
Many “experienced” students advise asking people to solve equations online. How to solve a trigonometric equation online, you ask. For online solutions tasks, you can go to forums on relevant topics, where they can help you with advice or in solving the problem. But it’s best to try to do it on your own.
Skills and abilities in solving trigonometric equations are very important and useful. Their development will require considerable effort from you. Many problems in physics, stereometry, etc. are associated with solving such equations. And the process of solving such problems itself presupposes the presence of skills and knowledge that can be acquired while studying the elements of trigonometry.
Learning trigonometric formulas
In the process of solving an equation, you may encounter the need to use any formula from trigonometry. You can, of course, start looking for it in your textbooks and cheat sheets. And if these formulas are stored in your head, you will not only save your nerves, but also make your task much easier, without wasting time searching necessary information. Thus, you will have the opportunity to think through the most rational way to solve the problem.