Bear in mind that, when it comes to trigonometric functions, the location of the exponent can be a bit misleading, however recall that sin²(θ) is really [ sin( θ )]²,
![\bf 2sin^2(2x)=2\implies sin^2(2x)=\cfrac{2}{2} \\\\\\ sin^2(2x)=1\implies [sin(2x)]^2=1\implies sin(2x)=\pm\sqrt{1} \\\\\\ sin(2x)=\pm 1\implies sin^{-1}[sin(2x)]=sin^{-1}(\pm 1)](https://tex.z-dn.net/?f=%5Cbf%202sin%5E2%282x%29%3D2%5Cimplies%20sin%5E2%282x%29%3D%5Ccfrac%7B2%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0Asin%5E2%282x%29%3D1%5Cimplies%20%5Bsin%282x%29%5D%5E2%3D1%5Cimplies%20sin%282x%29%3D%5Cpm%5Csqrt%7B1%7D%0A%5C%5C%5C%5C%5C%5C%0Asin%282x%29%3D%5Cpm%201%5Cimplies%20sin%5E%7B-1%7D%5Bsin%282x%29%5D%3Dsin%5E%7B-1%7D%28%5Cpm%201%29)
5x = 1/125 What is X
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Let z = sin(x). This means z^2 = (sin(x))^2 = sin^2(x). This allows us to go from the equation you're given to this equation: 7z^2 - 14z + 2 = -5
That turns into 7z^2 - 14z + 7 = 0 after adding 5 to both sides. Use the quadratic formula to solve for z. The only solution is z = 1 (see attached image). Since we made z = sin(x), this means sin(x) = 1. All solutions to this equation will be in the form x = (pi/2) + 2pi*n, which is the radian form of the solution set. If you need the degree form, then it would be x = 90 + 360*n
The 2pi*n (or 360*n) part ensures we get every angle coterminal to pi/2 radians (90 degrees), which captures the entire solution set.
Note: The variable n can be any integer.
