# How do you solve 2sin^3X-sin^2X-2SinX+1=0? I don't get how you get the 270.

## $2 {\sin}^{3} x - {\sin}^{2} x - 2 \sin x + 1 = 0$ for the interval $\left[0 , 2 \pi\right)$

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Sean Share
Feb 9, 2018

$x = {30}^{\circ} , {90}^{\circ} , {150}^{\circ} , {270}^{\circ}$

#### Explanation:

.

$2 {\sin}^{3} x - {\sin}^{2} x - 2 \sin x + 1 = 0$

Let's factor:

${\sin}^{2} x \left(2 \sin x - 1\right) - \left(2 \sin x - 1\right) = 0$

$\left({\sin}^{2} x - 1\right) \left(2 \sin x - 1\right) = 0$

Set each equal to zero and solve for $\sin x$

${\sin}^{2} x - 1 = 0$

${\sin}^{2} x = 1$

$\sin x = \pm 1 \therefore x = \frac{\pi}{2} \mathmr{and} {90}^{\circ} \mathmr{and} x = \frac{3 \pi}{2} \mathmr{and} {270}^{\circ}$

$2 \sin x - 1 = 0$

$2 \sin x = 1$

$\sin x = \frac{1}{2} \therefore x = \frac{\pi}{6} , \frac{5 \pi}{6} \mathmr{and} x = {30}^{\circ} , {150}^{\circ}$

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dk_ch Share
Feb 9, 2018

$2 {\sin}^{3} x - {\sin}^{2} x - 2 \sin x + 1 = 0$

$\implies {\sin}^{2} x \left(2 \sin x - 1\right) - 1 \left(2 \sin x - 1\right) = 0$

$\implies \left(2 \sin x - 1\right) \left({\sin}^{2} x - 1\right) = 0$

So
when $2 \sin x - 1$=0
$\sin x = \frac{1}{2} = \sin \left(\frac{\pi}{6}\right) = \sin \left(\pi - \frac{\pi}{6}\right)$

Hence $x = \frac{\pi}{6} , \frac{5 \pi}{6}$
when
${\sin}^{2} x - 1 = 0 \implies \sin x = \pm 1$

when
$\sin x = 1 = \sin \left(\frac{\pi}{2}\right)$
$\implies x = \frac{\pi}{2}$

Again

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

$\implies x = \frac{3 \pi}{2}$

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Feb 9, 2018

Factorize

#### Explanation:

$2 {\sin}^{3} x - {\sin}^{2} x - 2 \sin x + 1 = 0$
${\sin}^{2} x \left(2 \sin x - 1\right) - 1 \left(2 \sin x - 1\right) = 0$
$\left({\sin}^{2} x - 1\right) \left(2 \sin x - 1\right) = 0$
${\sin}^{2} x = 1$ or $2 \sin x = 1$
$\sin x = \pm 1$ or $\sin x = \frac{1}{2}$
$x = \frac{\pi}{2} , \frac{3 \pi}{2}$ or $x = \frac{\pi}{6} , 5 \frac{\pi}{6}$

$\frac{3 \pi}{2}$ is ${270}^{o}$

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