$2 \sin(x) - \sqrt{3} = 0$ $when\frac{\pi}{2} \leqslant x \leqslant \pi$

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Question

$2 \sin(x) - \sqrt{3} = 0$

$when\frac{\pi}{2} \leqslant x \leqslant \pi$

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Mathematics Thành Công 4 years 2021-07-26T17:06:50+00:00 2021-07-26T17:06:50+00:00 1 Answers 10 views 0

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Captcha* Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )

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thienhuong · Answer 1 · 2021-07-26T17:08:26+00:00

thienhuong

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2021-07-26T17:08:26+00:00 July 26, 2021 at 5:08 pm

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