How do you solve 2 sin x - 1 = 0 over the interval 0 to 2pi?

Apr 14, 2018

$x = \frac{\pi}{6} , 5 \frac{\pi}{6}$

Explanation:

1/ $2 \sin \left(x\right) - 1 = 0$
2/ $2 \sin \left(x\right) = 1$
3/ $\sin \left(x\right) = \frac{1}{2}$

4/ $x = \frac{\pi}{6} , 5 \frac{\pi}{6}$

Apr 14, 2018

$x = \frac{\pi}{6} \mathmr{and} \frac{5 \pi}{6}$

Explanation:

$2 \sin \left(x\right) - 1 = 0 | + 1$
2sin(x)=1|:2
$\sin \left(x\right) = \frac{1}{2}$
$x = \arcsin \left(\frac{1}{2}\right) = \frac{\pi}{6} \mathmr{and} \frac{5 \pi}{6}$

Apr 14, 2018

$x = \frac{\pi}{6} , \frac{5 \pi}{6}$

Explanation:

$2 \sin x - 1 = 0$

$\Rightarrow \sin x = \frac{1}{2}$

$\text{since "sinx>0" then x in first/second quadrant}$

$\Rightarrow x = {\sin}^{-} 1 \left(\frac{1}{2}\right) = \frac{\pi}{6} \leftarrow \textcolor{b l u e}{\text{first quadrant}}$

$\text{or "x=pi-pi/6=(5pi)/6larrcolor(blue)"second quadrant}$

$\Rightarrow x = \frac{\pi}{6} , \frac{5 \pi}{6} \to \left(0 , 2 \pi\right)$