Question

2 sin2x / ctgx + 3 cos2x = 1 - 2 cosx В ответах "нет корней".Я решал так:1) ОДЗ: ctgx ≠ 0 => x ≠ π/2 + πn, n ∈ Z2) 4 six2x + 3 cos2x = 1 - 2 cosx 4 (1 - cos2x) + 3 cos2x + 2 cos x - 1 = 0 - cos2x + 2 cos x + 3 = 0 | x (-1) cos2x - 2 cos x - 3 = 0 Замена: cos x = a, a ∈ [-1,1] a2 - 2a - 3 = 0 по т. Виета: а1 = 3, а2 = -1 Обратная замена: а) cosx = -1 x = π + 2πn, n ∈ Z б) cosx = 3 нет корней Ответ: x = π + 2πn, n ∈ ZНайдите, пожалуйста, ошибку.

Answer 1

это надо доказать тождество? или решить уравнение?

Answer 2

2sin2x/ctgx+3cos2x=1-2cosx;

2*sinx*cosx*sinx/cosx+3cos2x=1-2cosx;

2sin²x+3(cos²x-sin²x)=1-2cosx;

2-2cos²x+3cos²x-3+3cos²x=1-2cosx;

4cos²x+2cosx-2=0;

cos²x+(1/2)cosx-(1/2)=0;

cosx=-1;

x=π+2πn. n∈Z. - корень не подходит по ОДЗ;

cosx=1/2;

x=±π/3+2πn. n∈Z.

 

У меня так выходит.

 

Однаво у тебя ошибка в условии, ты квадраты написал так, что они означают аргументы угла...

А если это так, тогда корней действительно нет, ОДЗ: sinx≠0, cosx≠0, т.к. на ноль делить нельзя. x≠πn. x≠π/2+πn. Т.к. в условии котангенс, а в знаменателе выражения косинус.

Answer 3

Error in the Solution

In the given equation 2 sin^2(x) / ctg(x) + 3 cos^2(x) = 1 - 2 cos(x), you have correctly identified the domain of x as x ≠ π/2 + πn, where n ∈ Z. However, there seems to be an error in your solution. Let's go through the steps to find the mistake.

1. Step 1: Domain of x - You correctly identified the domain of x as x ≠ π/2 + πn, where n ∈ Z.

2. Step 2: Simplifying the Equation - You simplified the equation to 4 sin^2(x) + 3 cos^2(x) + 2 cos(x) - 1 = 0.

3. Step 3: Substitution - You made the substitution cos(x) = a, where a ∈ [-1, 1]. - The equation becomes 4(1 - a^2) + 3a^2 + 2a - 1 = 0.

4. Step 4: Solving the Quadratic Equation - You solved the quadratic equation 4 - 4a^2 + 3a^2 + 2a - 1 = 0. - Simplifying further, you obtained -a^2 + 2a + 3 = 0.

5. Step 5: Applying the Quadratic Formula - You applied the quadratic formula to solve for a. - The quadratic formula gives a = (-2 ± √(2^2 - 4(-1)(3))) / (2(-1)). - Simplifying further, you obtained a = (-2 ± √(4 + 12)) / (-2). - This simplifies to a = (-2 ± √16) / (-2). - Therefore, you found two possible values for a: a = 3 and a = -1.

6. Step 6: Inverse Substitution - You made the inverse substitution cos(x) = -1 and cos(x) = 3. - For cos(x) = -1, you obtained x = π + 2πn, where n ∈ Z. - For cos(x) = 3, you concluded that there are no solutions.

Identifying the Error

The error in your solution lies in the inverse substitution step. When you obtained the value cos(x) = 3, you concluded that there are no solutions. However, this is incorrect. The range of the cosine function is [-1, 1], which means that cos(x) cannot be greater than 1 or less than -1. Therefore, the equation cos(x) = 3 has no solutions.

Corrected Solution

To find the correct solution, we need to consider only the case where cos(x) = -1. Therefore, the correct solution to the equation 2 sin^2(x) / ctg(x) + 3 cos^2(x) = 1 - 2 cos(x) is x = π + 2πn, where n ∈ Z.

I hope this clarifies the error in your solution. Let me know if you have any further questions!