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?

Note that $\sin y=\sin x$ iff $\sin y-\sin x=0$ iff $$\sin\left(\frac{1}{2}(y-x)\right)\cos\left(\frac{1}{2}(y+x)\right)=0$$ iff 
$\sin\left(\frac{1}{2}(y-x)\right)=0$ or $\cos\left(\frac{1}{2}(y+x)\right)=0$ iff $y-x=2n\pi$ or $y+x=2(\pi/2+n\pi)$ iff
$y=x+2n\pi$ or $y=\pi-x+2n\pi$. From here, you should get your answer. When I work this out, your answer is actually correct.
An easier way is to factor and solve $\sin x=1/2$ and $\sin x=-1$. That way, you get
$x=-(1/2)\pi+2n\pi$ or $x=\pi/6+2n\pi$ or $x=5\pi/6+2n\pi$ for the answer. But this answer is equivalent to the above answer.
