Question

Number of solution of the equation ${\sin }^{-1}(4{\sin }^2\theta +\sin \theta )+{\cos }^{-1}(6\sin \theta -1)=\frac{\pi }{2}$ lying in the interval $[0,5\pi ],$ is-

• A. 9
• B. 7
• C. 6
• D. 5

Answer 1

The correct option is C 6

$\because$ ${\sin }^{-1}\alpha +{\cos }^{-1}\alpha =\frac{\pi }{2}$
$\Rightarrow$ $4{\sin }^2\theta +\sin \theta =6\sin \theta -1$
$\Rightarrow$ $4{\sin }^2-5\sin \theta +1=0$
$\Rightarrow$ $(4\sin \theta -1)(\sin \theta -1)=0$
$\Rightarrow$ $\sin \theta =\frac{1}{4}$ or $1$(reject)
$\because$ $-1\le 6\sin \theta -1\le 1$
$\Rightarrow$ $\sin \theta =\frac{1}{4}$
$\therefore$ No of solutions is $6.$ in $[0,5\pi ]$