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# 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 ${\text{A}}{\text{.}}\left( {\dfrac{\pi }{3}} \right) \\ {\text{B}}{\text{.}}\left( {\dfrac{\pi }{3},\dfrac{{5\pi }}{3}} \right) \\ {\text{C}}{\text{.}}\left( {\dfrac{\pi }{3},\dfrac{{5\pi }}{3},{{\cos }^{ - 1}}\left( {\dfrac{{ - 3}}{2}} \right)} \right) \\ {\text{D}}{\text{. None of these}}{\text{.}} \\$ Verified
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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 $\cos x = - \dfrac{3}{2}$
Now, as we know that range of a cosine function is [-1,1]
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 ${\text{B}}{\text{.}}\left( {\dfrac{\pi }{3},\dfrac{{5\pi }}{3}} \right)$

Note: Whenever we face such types of problems the key point to remember is that we need to have a good grasp over trigonometric properties, some of which have been used above. We must remember that the range of cosine is [-1,1]. This helps in getting us the required condition and gets us on the right track to reach the answer.
Last updated date: 26th Sep 2023
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