# Solve 2tan^2xsinx-tan^2x=0, if x in[0^@,360^@]?

Jan 13, 2018

$x \in \left({0}^{o} , {30}^{o} , {150}^{o} , {180}^{o} , {360}^{o}\right)$

#### Explanation:

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

$2 {\sin}^{2} \frac{x}{\cos} ^ 2 x \sin x - {\sin}^{2} \frac{x}{\cos} ^ 2 x = 0$

Multiply through by ${\cos}^{2} x$:

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

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

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

So either:

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

Solving for ${\sin}^{2} x = 0$

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

In the interval of ${0}^{o}$ to ${360}^{o}$ 3 possible values satisfy this, namely: ${0}^{o} , {180}^{o}$ and ${360}^{o}$.

Solving:

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

To values satisfy this in the interval, namely:

$x = {30}^{o}$ and $x = {180}^{o} - {30}^{o} = {150}^{o}$

So $x \in \left({0}^{o} , {30}^{o} , {150}^{o} , {180}^{o} , {360}^{o}\right)$