# Solve the following equation for all radian solutions and if 0\leq x\leq2\pi. Give all answers as exact values in radians. 0\leq x\leq2\pi

Solve the following equation for all radian solutions and if $0\le x\le 2\pi$. Give all answers as exact values in radians.
$0\le x\le 2\pi$
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Tasneem Almond

Given equation:
$2{\mathrm{sin}}^{2}x-\mathrm{sin}x-1=0$
$2{\mathrm{sin}}^{2}x-2\mathrm{sin}x+\mathrm{sin}x-1=0$
$2\mathrm{sin}x\left(\mathrm{sin}x-1\right)+1\left(\mathrm{sin}x-1\right)=0$
$\left(\mathrm{sin}x-1\right)\left(2\mathrm{sin}x+1\right)=0$
$\mathrm{sin}x-1=0$ or $2\mathrm{sin}x+1=0$
$\mathrm{sin}x=1$ or $2\mathrm{sin}x=-1$
$⇒\mathrm{sin}=\frac{-1}{2}$
$\mathrm{sin}x=1$
$x=2\pi +\frac{\pi }{2}$, where is any integer
$\mathrm{sin}x$ is negative in quadrant 3 and 4.
Then general solution is
$\mathrm{sin}x=\frac{-1}{2}$
$⇒x=\frac{7\pi }{6}+2k\pi ,\frac{11\pi }{6}+2k\pi$ for any inter
In $0\le x<2\pi$, put x=0 we get
$x=\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6}$

Jeffrey Jordon