Why is this solution to $2\sin^2(x) + \sin(x) - 1 = 0$ incorrect? $$2\sin^2(x) + \sin(x) - 1 = 0$$
$$\sin(x) = \cos (2x)$$
$$ \sin(x) = \sin (2x  + \dfrac{1}{2} \pi)$$
$$ x = -\dfrac{1}{2}\pi - k \cdot 2\pi$$ or $$ x = \dfrac{1}{6} \pi + k \cdot \dfrac{2}{3} \pi$$
Why is this incorrect?

Recall that
$$\sin(A) = \sin(B) \implies A = n \pi +(-1)^n B$$
Also, you could solve your problem in a slightly easier way. Note that
$$2 \sin^2(x) + \sin(x) - 1 = (2 \sin(x)-1)(\sin(x)+1) = 0 \implies \sin(x) = \dfrac12 \text{ or }-1$$
Hence,
$$x = n \pi +(-1)^n \dfrac{\pi}6, 2n \pi - \dfrac{\pi}2$$
