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

Answer: (A)

$9$

$4si{n}^2\theta +sin\theta =6sin\theta -14si{n}^2\theta -5sin\theta +1=04si{n}^2\theta -4sin\theta -sin\theta +1=04sin\theta (sin\theta -1)-1(sin\theta -1)=0(4sin\theta -1)(sin\theta -1)=0\therefore sin\theta =\left(\frac{1}{4}\right)orsin\theta =1in\theta \in [0,5\pi ]forsin\theta =1,\theta =3solutionsforsin\theta =\left(\frac{1}{4}\right),\theta =6solutionshencetotal=9solutions$