We have:
[tex]\cos 2x = 1 - 2 \sin^2 x[/tex]
Thus:
[tex]1 - 2 \sin^2 x = \sin x[/tex]
Adding [tex]2 \sin^2 x - 1[/tex] gives:
[tex]2 \sin^2 x + \sin x - 1 = 0[/tex]
Factoring gives:
[tex](2 \sin x + 1)(\sin x - 1) = 0[/tex]
Thus, we have:
[tex]\sin x = \frac{-1}{2}[/tex]
or
[tex]\sin x = 1[/tex].
Note that because [tex]2 \pi > 2[/tex], there will be no solutions [tex]y[/tex] of the form [tex]2n\pi + z[/tex] for positive integer [tex]n[/tex] and solution [tex]z[/tex]. So we find the basic values of [tex]x[/tex] giving [tex]\sin x \in (\frac{-1}{2}, 1)[/tex].
For [tex]\sin x = \frac{-1}{2}[/tex] with positive [TEX]x[/tex], [tex]x > \pi > 2[/tex] (this can be seen easily from the unit circle. But [tex]x < 2[/tex], a contradiction. So we examine the case in which [tex]\sin x = 1[/tex].
In this case, [tex]x = \frac{\pi}{2}[/tex]. So this is the only solution to this equation.
Substituting this back in, we have [tex]\sin \frac{\pi}{2} = \cos \pi = 1[/tex], as desired. So this is a valid solution.
Thus, [tex]x = \frac{\pi}{2}[/tex].
