Question

= 0" alt="2 \sin(x) - \sqrt{3} = 0" align="absmiddle" class="latex-formula">
$when\frac{\pi}{2} \leqslant x \leqslant \pi$
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Answer 1

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 $\dfrac{2\pi}{3}$.