# What are all the solutions between 0 and 2π for 2cos^2 = sinx+1?

Mar 29, 2016

$x \in \left\{\frac{\pi}{6} , \frac{5 \pi}{6} , \frac{3 \pi}{2}\right\}$

#### Explanation:

Assuming the equation was meant to be:
$\textcolor{w h i t e}{\text{XXX}} 2 {\cos}^{2} \left(\textcolor{red}{x}\right) = \sin \left(x\right) + 1$

Using the relationship:
$\textcolor{w h i t e}{\text{XXX}} {\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$
$2 {\cos}^{2} \left(x\right) = \sin \left(x\right) + 1$

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

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

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

$\rightarrow \sin \left(x\right) = \frac{1}{2} \textcolor{w h i t e}{\text{XXX")orcolor(white)("XXX}} \sin \left(x\right) = - 1$
$\rightarrow x = \frac{\pi}{6} \mathmr{and} \frac{5 \pi}{6} \textcolor{w h i t e}{\text{XXXXXXX}} \rightarrow x = \frac{3 \pi}{2}$
$\textcolor{w h i t e}{\text{XXXXXXXXXX}}$(for $x \in \left[0 , 2 \pi\right]$)