 # 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 sanuluy 2021-09-12 Answered
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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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
$x=\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6}$

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