Question

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

Answer 1

Answer:

$x=0, \dfrac{\pi}{6},\dfrac{5\pi}{6},\pi \:\sf(for\:the\:given\:interval)$

Step-by-step 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:

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

$\sin(x)=0$

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

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

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

$\implies \sin(x)=\dfrac12$

$\implies x=\dfrac{\pi}{6} \pm2 \pi n, \dfrac{5\pi}{6} \pm2 \pi n$

$\implies x=\dfrac{\pi}{6},\dfrac{5\pi}{6}\:\sf(for\:the\:given\:interval)$

Final solution:

$x=0, \dfrac{\pi}{6},\dfrac{5\pi}{6},\pi \:\sf(for\:the\:given\:interval)$

Answer 2

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