2sin3x/2 cos3x/2 -sin^2 3x=0

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вопрос

To solve this trigonometric equation, we can simplify it by breaking it down into smaller components using trigonometric identities.

Given: 2sin(3x/2)cos(3x/2) - sin^2(3x) = 0

Use the double angle formula for sine: sin(2θ) = 2sin(θ)cos(θ)

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

So, the equation becomes: sin(3x) - sin^2(3x) = 0

Now, we can use the Pythagorean identity: sin^2θ + cos^2θ = 1

Rearranging it, we get: sin^2θ = 1 - cos^2θ

Plugging this into the equation, we get: sin(3x) - (1 - cos^2(3x)) = 0

Simplify further: sin(3x) - 1 + cos^2(3x) = 0

Now, use the double angle formula for cosine: cos(2θ) = 1 - 2sin^2(θ)

cos(2θ) = cos^2(3x) - sin^2(3x)

Substitute into the equation: sin(3x) - 1 + cos(2(3x)) = 0

sin(3x) - 1 + cos(6x) = 0

There is no simple way to solve this equation further without using numerical methods or specific techniques for trigonometric equations.