QJEEM22Maba27: JEE Main 2022 Mathematics - Class 11 JEEM22Maba27 Solution: The number of solutions of the equation in the interval (0, 10) …

$\sin x = {\cos ^2}x$ , $x \in (0,10)$

$\Rightarrow \sin x = 1 - {\sin ^2}x$

$\Rightarrow {\sin ^2}x + \sin x - 1 = 0$

$\therefore$ $\sin x = {{ - 1 \pm \,\sqrt {1 + 4} } \over 2}$

$\Rightarrow \sin x = {{ - 1 \pm \,\sqrt 5 } \over 2}$

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