# How to solve 2cos^(2)x-sinx=1 ?

## I do understand how I get sinx which is sinx=1/2 and sinx=-1 but I do not understand what I should do next. I know that I should look at the unit circle and when I do that I get 3pi/2, pi/6 and 5pi/6 but the answer is x=pi/6+n2pi/3

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Oct 24, 2017

$x = n \pi + {\left(- 1\right)}^{n} \frac{\pi}{6}$ or $x = 2 n \pi - \frac{\pi}{2}$

#### Explanation:

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

$\therefore 2 {\cos}^{2} x - \sin x = 2 - 2 {\sin}^{2} x - \sin x = 1$

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

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

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

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

$\sin x = \frac{1}{2} , - 1$

If $\sin x = \frac{1}{2} = \sin \left(\frac{\pi}{6}\right)$ and $x = n \pi + {\left(- 1\right)}^{n} \frac{\pi}{6}$ and in $\left[0 , 2 \pi\right)$ $x = \frac{\pi}{6}$ or $\frac{5 \pi}{6}$

and if $x = - 1 = \sin \left(- \frac{\pi}{2}\right)$ and $x = 2 n \pi - \frac{\pi}{2}$ and in $\left[0 , 2 \pi\right)$ $\frac{3 \pi}{2}$

Here $n$ is an integer.

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