Respuesta :

loneguy

[tex]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[/tex]

semsee45

Answer:

[tex]x=0, \dfrac{\pi}{6},\dfrac{5\pi}{6},\pi \:\sf(for\:the\:given\:interval)[/tex]

Step-by-step explanation:

given interval [0, 2π) = 0 ≤ x < 2π

[tex]2\sin^2(x)=\sin(x)[/tex]

subtract sin(x) from both sides:

[tex]\implies 2\sin^2(x)-\sin(x)=0[/tex]

factor:

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

[tex]\sin(x)=0[/tex]

[tex]\implies x=0 \pm2 \pi n, \pi \pm2 \pi n[/tex]

[tex]\implies x=0, \pi \:\sf(for\:the\:given\:interval)[/tex]

[tex]2\sin(x)-1=0[/tex]

[tex]\implies \sin(x)=\dfrac12[/tex]

[tex]\implies x=\dfrac{\pi}{6} \pm2 \pi n, \dfrac{5\pi}{6} \pm2 \pi n[/tex]

[tex]\implies x=\dfrac{\pi}{6},\dfrac{5\pi}{6}\:\sf(for\:the\:given\:interval)[/tex]

Final solution:

[tex]x=0, \dfrac{\pi}{6},\dfrac{5\pi}{6},\pi \:\sf(for\:the\:given\:interval)[/tex]

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