# How do you solve 2sin^2x-sinx=1?

Add a $- \sin x$ to both sides to get

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

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

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

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

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

From the last equation we have

$\sin x = 1 \implies x = \frac{1}{2} \left(4 \cdot \pi \cdot n + \pi\right)$ where $n$ integer

and

$\sin x = - \frac{1}{2}$ with solutions

${x}_{1} = \frac{1}{6} \left(12 \cdot \pi \cdot k - \pi\right)$

and

${x}_{2} = \frac{1}{6} \left(12 \cdot \pi \cdot k + 7 \cdot \pi\right)$

where $k$ integer

Oct 31, 2016

$x = \left(\frac{7 \pi}{6} + 2 n \pi\right)$, $x = \left(\frac{11 \pi}{6} + 2 n \pi\right)$ and $x = \left(\frac{\pi}{2} + 2 n \pi\right)$ where $n \in \mathbb{N}$.

#### Explanation:

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

Subtract $1$ from each side.#

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

Factor

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

Set each factor equal to zero and solve.

$2 \sin x + 1 = 0$ and $\sin x - 1 = 0$

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

For $\sin x = - \frac{1}{2}$ the unit circle shows that $x = \frac{7 \pi}{6}$ and $x = \frac{11 \pi}{6}$ within $0 \le x < 2 \pi$

$\implies x = \frac{7 \pi}{6} + 2 n \pi$ and $x = \frac{11 \pi}{6} + 2 n \pi$ where $n$ is an integer.

For $\sin x = 1$, the unit circle shows that $x = \frac{\pi}{2}$ for $0 \le x < 2 \pi$

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