Sin^2 2x=sin^2 и
2sin2x*sin6x=cos4x

Предыдущий
вопрос Следующий
вопрос

To solve for x in the equations given, we first need to rewrite the equations using trigonometric identities.

Using the double angle formula, sin 2x = 2sin x cos x. Therefore, (2sin x cos x)^2 = sin^2 x, which simplifies to 4sin^2 x cos^2 x = sin^2 x.

Taking the square root of both sides, we get:

2sin x cos x = sin x

Divide by sin x:

2cos x = 1

Cos x = 1/2

x = π/3 + 2kπ or x = 5π/3 + 2kπ, where k is an integer.

Using the double angle formula, sin 2x = 2sin x cos x and sin 6x = 2sin 3x cos 3x. Therefore, the equation becomes:

2(2sin x cos x)(2sin 3x cos 3x) = cos 4x

Simplify:

8sin x cos x sin 3x cos 3x = cos 4x

Apply trigonometric identities sin 3x = 3sin x - 4sin^3 x and cos 3x = 4cos^3 x - 3cos x:

8sin x cos x(3sin x - 4sin^3 x)(4cos^3 x - 3cos x) = cos 4x

Simplify further and solve for x.

Note: The solutions for x can be many due to the periodic nature of trigonometric functions.