Question
The number of values of $$x$$ in the interval $$\left[ {0,3\pi } \right]$$ satisfying the equation $$2{\sin ^2}x + 5\sin x - 3 = 0$$ is
A.
4
B.
6
C.
1
D.
2
Answer :
4
Solution :
$$\eqalign{ & 2{\sin ^2}x + 5\sin x - 3 = 0 \cr & \Rightarrow \,\,\left( {\sin x + 3} \right)\left( {2\sin x - 1} \right) = 0 \cr & \Rightarrow \,\,\sin x = \frac{1}{2}\,\,\,{\text{and }}\sin x \ne - 3 \cr & \therefore \,\,{\text{In }}\left[ {0,3\pi } \right],x\,\,{\text{has 4 values}}{\text{.}} \cr} $$
$$\eqalign{ & 2{\sin ^2}x + 5\sin x - 3 = 0 \cr & \Rightarrow \,\,\left( {\sin x + 3} \right)\left( {2\sin x - 1} \right) = 0 \cr & \Rightarrow \,\,\sin x = \frac{1}{2}\,\,\,{\text{and }}\sin x \ne - 3 \cr & \therefore \,\,{\text{In }}\left[ {0,3\pi } \right],x\,\,{\text{has 4 values}}{\text{.}} \cr} $$