# How do you solve 2sin^2x-5sinx+2=0?

Oct 4, 2016

$x = \left\{\frac{\pi}{6}\right\}$

#### Explanation:

Use the substitution method and then factor

Let $u = \sin x$

so, ${u}^{2} = {\sin}^{2} x$

$2 \textcolor{red}{{u}^{2}} - 5 \textcolor{red}{u} + 2 = 0$

Now you can factor

Multiply the coefficient of the first term, $2$, with the last term, $2$.

$2 \cdot 2 = 4$

Ask yourself what are the factors of $4$ that add up to the coefficient of the middle term, $- 5$?

Factors of 4

$\left(1\right) \left(4\right)$ $\implies N O$
$\textcolor{red}{\left(- 1\right) \left(- 4\right)}$ $\implies \textcolor{red}{Y E S}$
$\left(2\right) \left(2\right)$ $\implies N O$
$\left(- 2\right) \left(- 2\right)$ $\implies N O$

Now place those factors in an order that makes it easy to factor by grouping.

$\left(2 {u}^{2} \textcolor{red}{- 4 u}\right) + \left(\textcolor{red}{- 1 u} + 2\right) = 0$

Factor out $2 u$ from the first grouping.

Factor out $- 1$ from the second grouping.

$\textcolor{red}{2 u} \left(u - 2\right) \textcolor{red}{- 1} \left(u - 2\right) = 0$

Now you can factor out a grouping, $\left(u - 2\right)$

$\left(u - 2\right) \left(2 u - 1\right) = 0$

Now use the Zero Property

$u - 2 = 0$ and $2 u - 1 = 0$

$u = 2$ and $u = \frac{1}{2}$

Now switch back to $\sin x$

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

$\cancel{\sin x = 2}$ is discarded because $\sin x$ oscillates between -1 and 1.

Falling back to trigonometry $\sin x$ is in the ratio $\frac{y}{r}$

$\frac{y}{r} = \frac{1}{2}$ which means that $x = \sqrt{3}$ by using the pythagorean theorem $x = \sqrt{{2}^{2} - {1}^{2}} = \sqrt{4 - 1} = \sqrt{3}$ or knowing that we have the special triangle , $30 , 60 , 90$ which corresponds to the sides $1 , \sqrt{3} , 2.$

The side of length $1$ corresponds to $30$ degrees which is also $\frac{\pi}{6}$

All of that to say that

$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

$x = \left\{\frac{\pi}{6}\right\}$

I have tutorials on methods of factoring found here, https://www.youtube.com/playlist?list=PLsX0tNIJwRTxqIIcfpsTII9_xhqVIbQlp