Question

Помогите решить 1-2sin^2x=2cos2x Варианты ответа: 1) (пи)/4+(пи)k/2, k принадлежит Z 2)(пи)/2+(пи)k, k принадлежит Z 3) Другое решение Спасибо!

Answer 1

1-2sin²x=2cos2x
1-2sin²x=2(cos²x-sin²x),  1-2sin²x=2cos²x-2sin²x
2cos²x-2sin²x+2sin²x=1, 2cos²x=1, cos²x=1/2,
cosx=1/√2   и  cosx=-1/√2
x=+-π/4+2πk    x= +-3π/4+2πk или
х=π/4+πk/2,k∈Z

Answer 2

I can help you with your math problem. Here is a possible solution:

To solve the equation $1-2\sin^2 x = 2\cos 2x$, we can use the double angle identity $\cos 2x = 1-2\sin^2 x$. Then we get:

$$1-2\sin^2 x = 1-2\sin^2 x$$

This is an identity, which means it is true for all values of $x$. Therefore, the solution set is the set of all real numbers, or $(-\infty, \infty)$.

Alternatively, we can use another double angle identity $\cos 2x = \cos^2 x - \sin^2 x$. Then we get:

$$1-2\sin^2 x = \cos^2 x - \sin^2 x$$

Adding $2\sin^2 x$ to both sides, we get:

$$1 = \cos^2 x + \sin^2 x$$

This is also an identity, which means it is true for all values of $x$. Therefore, the solution set is the same as before, $(-\infty, \infty)$.

You can check your answer using online calculators such as [Mathway](https://www.mathway.com/trigonometry) or [Symbolab](https://www.symbolab.com/solver/trigonometric-equation-calculator). I hope this helps. Have a nice day!