$\sin x = {\cos ^2}x$, $x \in (0,10)$ $\Rightarrow \sin x = 1 - {\sin ^2}x$ $\Rightarrow {\sin ^2}x + \sin x - 1 = 0$ $\therefore$ $\sin x = {{ - 1 \pm \,\sqrt {1 + 4} } \over 2}$ $\Rightarrow \sin x = {{ - 1 \pm \,\sqrt 5 } \over 2}$ We know $\sin \in ( - 1,1)$ $\therefore$ ${{ - 1 - \sqrt 5 } \over 2}$ can't be a value of sin x $\therefore$ $\sin x = {{\sqrt 5 - 1} \over 2}$ $3\pi = 3 \times 3.14 = 9.42 < 10$ ${{7\pi } \over 2} = {7 \over 2} \times 3.14 = 10.99 > 10$ $\therefore$ 10 will be in between 3$\pi$ and ${{7\pi } \over 2}$. There are 4 intersection at A, B, C and D between sin x graph and $y = {{\sqrt 5 - 1} \over 2}$ graph. $\therefore$ possible solution = 4