Question

# The solution set of $\left( {2\cos x - 1} \right)\left( {3 + 2\cos x} \right) = 0$ in the interval $0 \leqslant x \leqslant 2\pi$ is ${\text{A}}{\text{.}}\left( {\dfrac{\pi }{3}} \right) \\ {\text{B}}{\text{.}}\left( {\dfrac{\pi }{3},\dfrac{{5\pi }}{3}} \right) \\ {\text{C}}{\text{.}}\left( {\dfrac{\pi }{3},\dfrac{{5\pi }}{3},{{\cos }^{ - 1}}\left( {\dfrac{{ - 3}}{2}} \right)} \right) \\ {\text{D}}{\text{. None of these}}{\text{.}} \\$

Hint: In this question we have to find the solution set of values of x. We would be using the fact that if two numbers are in multiplication and equal to zero then, one of them must be equal to zero. Using this we will be able to reach the desired answer.

We have been given the equation, $\left( {2\cos x - 1} \right)\left( {3 + 2\cos x} \right) = 0$
So, either $2\cos x - 1 = 0$ or $3 + 2\cos x = 0$
$\Rightarrow \cos x = \dfrac{1}{2}$ or $\cos x = - \dfrac{3}{2}$
So, $\cos x \ne - \dfrac{3}{2}$
$\Rightarrow \cos x = \dfrac{1}{2}$
$\Rightarrow x = {\cos ^{ - 1}}\dfrac{1}{2}$
$\Rightarrow x = \dfrac{\pi }{3},\dfrac{{5\pi }}{3}$
Hence, the correct answer is ${\text{B}}{\text{.}}\left( {\dfrac{\pi }{3},\dfrac{{5\pi }}{3}} \right)$