Question

Find all solutions in the interval [0,2pi).
2sin^2(x)=sin(x)

Answer 1

`2\sin^2 x =\sin x \\\\\implies 2 \sin^2 x - \sin x=0\\\\\implies \sin x(2 \sin x -1) =0\\\\\implies \sin x = 0~~\text{or}~~ 2 \sin x -1 =0\\\\\implies \sin x = 0~~ \text{or}~ ~ \sin x = \frac 12\\\\\text{For}~~ \sin x = 0\\\\x=n\pi \\\\\text{In the interval,}~~[0, 2\pi)\\\\x=0, \pi\\\\\text{For}~~ \sin x = \frac 12\\\\x=n\pi+(-1)^n \frac{\pi}6 \\\\\text{In the interval,}~~[0, 2\pi)\\\\x=\frac{\pi}6,~ \frac{5\pi}6\\\\\text{Hence,}~ x = 0,~\pi,~ \frac{\pi}6, ~\frac{5 \pi}6`

Answer 2

`x=0, (\pi)/(6),(5\pi)/(6),\pi \:\sf(for\:the\:given\:interval)`

Explanation:

given interval [0, 2π) = 0 ≤ x < 2π

`2\sin^2(x)=\sin(x)`

subtract sin(x) from both sides:

`\implies 2\sin^2(x)-\sin(x)=0`

factor:

`\implies \sin(x)[2\sin(x)-1]=0`

`\sin(x)=0`

`\implies x=0 \pm2 \pi n, \pi \pm2 \pi n`

`\implies x=0, \pi \:\sf(for\:the\:given\:interval)`

`2\sin(x)-1=0`

`\implies \sin(x)=\frac12`

`\implies x=(\pi)/(6) \pm2 \pi n, (5\pi)/(6) \pm2 \pi n`

`\implies x=(\pi)/(6),(5\pi)/(6)\:\sf(for\:the\:given\:interval)`

Final solution:

`x=0, (\pi)/(6),(5\pi)/(6),\pi \:\sf(for\:the\:given\:interval)`