# What are the general solution of equation 2sin^2x-sinx-1=0?

Aug 1, 2016

Roots are given by $x = n \pi - {\left(- 1\right)}^{n} \left(\frac{\pi}{6}\right\}$

and $x = 2 n \pi + \frac{\pi}{2}$, where $n$ is an integer

#### Explanation:

Roots of $2 {\sin}^{2} x - \sin x - 1 = 0$ can be obtained by factorizing the function.

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

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

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

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

and roots of $2 {\sin}^{2} x - \sin x - 1 = 0$ are given by

Hence, either $2 \sin x + 1 = 0$ i.e. $\sin x = - \frac{1}{2}$ and

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

or $\sin x - 1 = 0$ i.e. $\sin x = 1$ and

$x = 2 n \pi + \frac{\pi}{2}$, where $n$ is an integer