341If \cos x \ge 0, then we have
\sin x + \cos x \ge \sin^2 x + \cos^2 x =1$
Hence\sin x + \cos x+\sqrt{\sin x} + \sin^2 x \ge 1$ and
equality occurs only when \sin x = 0 i.e. x = 2k\pi, k \in \mathbb{Z}
If \cos x<0 then \cos x =- \sqrt{1-\sin^2 x} and
the equation may be written as
\sqrt{\sin x} + \sin x = (1-\sin^2 x) + \sqrt{1-\sin^2 x}
The function f(x) = y+y^2 is monotonic when y>0
Hence, we must have \sin x = 1-\sin^2x which yields
\sin x = \frac{\sqrt{5}-1}{2}.
Since we have cos x<0, the solutions are all given by
x = \pi - \sin^{-1} \left(\frac{\sqrt{5}-1}{2}\right) + 2k\pi, k \in \mathbb{Z}
In this way all possible solutions have been obtained
