A geometry problem by Aly Ahmed

Of all the solutions to the equation after removing the square roots, only $x=\dfrac{π}{2}$ satisfies the given equation. So, this is the principal root. The general solution is $\boxed {x=\dfrac{π}{2}+2kπ}$ , where $k$ is an integer.

@Aly Ahmed , this problem should not be set as an objective question. Because just input the four options and everyone gets the answer and no one learns about Geometry.

My Geometry solution as follows. Let's consider $0 \le x < 2\pi$ . We note that the LHS, $\sqrt{\sin x-\sqrt{\sin x + \cos x}} \ge 0$ , then the RHS, $\cos x \ge 0$ , $\implies 0 \le x \le \frac \pi 2$ .

Now, for the LHS $\ge 0$ , $\sin x \ge \sqrt{\sin x + \cos x} \implies \sin^2 x \ge \sin x + \cos x = \sqrt 2 \sin \left(x+\frac \pi 4\right)$ . For $0 \le x \le \frac \pi 2$ , we note that $\sin^2 x \le 1$ , but $\sqrt 2 \sin \left(x + \frac \pi 4\right) \ge 1$ . There is only one solution, that is $\sin^2 x = \sqrt 2 \sin \left(x + \frac \pi 4\right) = \sin x + \cos x = 1$ or $x = \frac \pi 2$ .

Therefore, the general solution of $x = \boxed{\frac \pi 2 + 2k\pi}$ , where $k$ is an integer.