# sin2x cosx=0

#### Explanation:

We know that $\sin 2 x = 2 \sin x \cos x$

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So this becomes,

$2 \sin x \cos x + \cos x = 0$

$\left(2 \sin x + 1\right) \left(\cos x\right) = 0$

So either $2 \sin x + 1 = 0$ so sánh $\sin x = - \frac{1}{2}$ in which case $x = \frac{7 \pi}{6}$
Or, $\cos x = 0$ in which case $x = \frac{\pi}{2}$

So $x = \frac{\pi}{2} , \frac{7 \pi}{6}$

#### Explanation:

$S \in \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$

$2 \sin \left(x\right) \cos \left(x\right) = - \cos \left(x\right)$

Xem thêm: tiếng anh lớp 7 trang 120

$\sin \left(x\right) = - \frac{1}{2}$

${\sin}^{-} 1 \left(x\right) = x$

sin of x is negative value Only at the third and the forth quarter.

$x = 210 \mathmr{and} - 30 \mathmr{and} 330$

# pi/2; (7pi)/6; (3pi)/2; (11pi)/6# for interval $\left(0 , 2 \pi\right)$

Xem thêm: soạn mây và sóng lớp 7

#### Explanation:

sin 2x + cos x = 0
2sin x.cos x + cos x = 0
cos x(2sin x + 1) = 0
either factor should be zero.
a. cos x = 0
Unit circle gives 2 solutions -->
$x = \frac{\pi}{2} + 2 k \pi$, and
$x = \frac{3 \pi}{2} + 2 k \pi$.
b. 2sin x + 1 = 0 --> $\sin x = - \frac{1}{2}$
Trig table and unit circle give 2 solutions:
$x = - \frac{\pi}{6} + 2 k \pi$, or $x = \frac{11 \pi}{6} + 2 k \pi$ (co-terminal)
$x = \pi - \left(- \frac{\pi}{6}\right) = \frac{7 \pi}{6} + 2 k \pi$

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