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

User Guapolo
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2 Answers

17 votes
17 votes


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

User Chase Henslee
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26 votes
26 votes

Answer:


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)

Find all solutions in the interval [0,2pi). 2sin^2(x)=sin(x)-example-1
User FatherFigure
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