Question

Find all solutions in the interval [0, 2π) for 2 sin^2 x = sin x






Option A: x = pi divided by three., two pi divided by three.

Option B: x = pi divided by two, three pi divided by two., pi divided by three., two pi divided by three.

Option C: x = 0, π, pi divided by six, five pi divided by six

Option D: x = pi divided by six, five pi divided by six

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Answer 1

Answer:

D

Step-by-step explanation:

2 sin²x=sin x

2 sin² x-sin x=0

sin x(2 sin x-1)=0

sin x=0

x=0,π

2 sin x-1=0

2 sin x=1

sin x=1/2= sin π/6,sin (π-π/6)

x=π/6,5π/6