This question is similar in form to this one:
Finding all Trigonometric Solutions of an Equation within a Given Interval
However, I want to verify my method of solving, as it would appear I have made some logical error in my processes.
My method goes as follows:
Solve for $\theta$ over the interval $[0, 2\pi)$
$\sqrt{2}\sin{\theta} = \sqrt{\sin{\theta}+1}$
$(\sqrt{2}\sin{\theta})^2 = (\sqrt{\sin{\theta}+1})^2$
$2\sin^2{\theta} = \sin{\theta} + 1$
$\sin^2{\theta} = \dfrac{\sin{\theta} + 1}{2}$
$\dfrac{1-\cos{2\theta}}{2} = \dfrac{\sin{\theta} + 1}{2}$
$1-\cos{2\theta} = \sin{\theta} + 1$
$-\cos{2\theta} = \sin{\theta}$
$0 = \sin{\theta} + \cos{2\theta}$
Then solve for $\theta$.
$\sin{\dfrac{1\pi}{2}} = 1$
$\cos{\dfrac{2\pi}{2}} = \cos{\pi} = -1$
Therefore $\theta = \dfrac{\pi}{2}$
I was told my method was incorrect. I don't see where.