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igomit [66]
3 years ago
9

= 0" alt="2 \sin(x) - \sqrt{3} = 0" align="absmiddle" class="latex-formula">
when\frac{\pi}{2} \leqslant x \leqslant  \pi
​
Mathematics
1 answer:
atroni [7]3 years ago
6 0

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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4 0
2 years ago
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4 years ago
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