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) = 02(-sin(3π/2 + x))^2 + sin(2x) = 02sin^2(π/2 - x) + 2sin(x)cos(x) = 02cos^2(x) + 2sin(x)cos(x) = 02cos(x)(cos(x) + sin(x)) = 0cos(x)(cos(x) + sin(x)) = 0
Now, there are two cases that would satisfy this equation:
cos(x) = 0This would occur when x = π/2 + nπ where n is an integer.
cos(x) + sin(x) = 0Recall that cos(x) = sin(π/2 - x)So, sin(π/2 - x) + sin(x) = 0sin(π/2 - x) = -sin(x)cos(x) = -sin(x)tan(x) = -1x = 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.
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.