Given:
The equation is
[tex]2\sin x-\sqrt{3}=0[/tex]
When [tex]\dfrac{\pi}{2}\leq x\leq \pi[/tex].
To find:
The value of x.
Solution:
We have,
[tex]2\sin x-\sqrt{3}=0[/tex]
It can be written as
[tex]2\sin x=\sqrt{3}[/tex]
[tex]\sin x=\dfrac{\sqrt{3}}{2}[/tex]
It is given that [tex]\dfrac{\pi}{2}\leq x\leq \pi[/tex].
[tex]\sin x=\sin \left(\pi-\dfrac{\pi}{3}\right)[/tex]
[tex]x=\dfrac{3\pi-\pi}{3}[/tex]
[tex]x=\dfrac{2\pi}{3}[/tex]
Therefore, the value of x is [tex]\dfrac{2\pi}{3}[/tex].
