JEE Advance - Mathematics (1995 - No. 21)
The general values of $$\theta $$ satisfying the equation $$2{\sin ^2}\theta - 3\sin \theta - 2 = 0$$ is
$$n\pi + {\left( { - 1} \right)^n}\pi /6$$
$$n\pi + {\left( { - 1} \right)^n}\pi /2 $$
$$n\pi + {\left( { - 1} \right)^n}5\pi /6$$
$$n\pi + {\left( { - 1} \right)^n}7\pi /6$$
Comments (0)
