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    2021-07-26T17:08:26+00:00

    Given:

    The equation is

    2\sin x-\sqrt{3}=0

    When \dfrac{\pi}{2}\leq x\leq \pi.

    To find:

    The value of x.

    Solution:

    We have,

    2\sin x-\sqrt{3}=0

    It can be written as

    2\sin x=\sqrt{3}

    \sin x=\dfrac{\sqrt{3}}{2}

    It is given that \dfrac{\pi}{2}\leq x\leq \pi.

    \sin x=\sin \left(\pi-\dfrac{\pi}{3}\right)

    x=\dfrac{3\pi-\pi}{3}

    x=\dfrac{2\pi}{3}

    Therefore, the value of x is \dfrac{2\pi}{3}.

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