2сos^2(3pi/2 + x) + sin2x=0

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To simplify the given expression 2cos^2(3pi/2 + x) + sin2x = 0, we can use the trigonometric identities to rewrite the cosine and sine functions in terms of sine functions.

Recall that cos(π/2 + θ) = -sin(θ) and sin(2θ) = 2sin(θ)cos(θ).

Applying these identities, we have:

2cos^2(3π/2 + x) + sin(2x) = 0
2(-sin(3π/2 + x))^2 + sin(2x) = 0
2sin^2(π/2 - x) + 2sin(x)cos(x) = 0
2cos^2(x) + 2sin(x)cos(x) = 0
2cos(x)(cos(x) + sin(x)) = 0
cos(x)(cos(x) + sin(x)) = 0

Now, there are two cases that would satisfy this equation:

cos(x) = 0
This would occur when x = π/2 + nπ where n is an integer.

cos(x) + sin(x) = 0
Recall that cos(x) = sin(π/2 - x)
So, sin(π/2 - x) + sin(x) = 0
sin(π/2 - x) = -sin(x)
cos(x) = -sin(x)
tan(x) = -1
x = 3π/4 + nπ where n is an integer.

Therefore, the solutions to the equation 2cos^2(3π/2 + x) + sin(2x) = 0 are x = π/2 + nπ, 3π/4 + nπ where n is an integer.