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marasofia1616
Find all solutions in the interval [0, 2π). 2 sin2^x = sin x
\[2\sin 2^{x}=\sin x \] are you sure it looks like that?
\[2 \sin ^{2}x=\sin x\]
sorry this is how it looks like
pi / 4 = 45 5pi / 4 = 135
2 sin^2 x = sin x 2 sin x = 1 dividing both sides by sin x sin x = 1 / 2 dividing both sides by 2 arcsin (1/2) = x so x = pi/4, 5pi/4