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Find all solutions in the interval [0, 2π).

0 votes

2sin^2 x = sin x

asked Jun 18, 2013 in TRIGONOMETRY by andrew Scholar

1 Answer

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2sin^2x  = sinx

2sin^2x  - sinx  = 0

sinx(2sinx - 1) = 0

2sinx - 1 = 0

2sinx  = 1

sinx = 1/2

sinx = 30º

The  solution is x = 0º,π/6, π

answered Jun 18, 2013 by anonymous

sin(x) = 0 and sin(x) = 1/2.

sin(x) = sin(0) and sin(x) = sin(π/6).

General solution : If sin θ = sin α, then θ = nπ + (- 1)n α, where n is an integer.

If α = 0 then θ = nπ.

If α = 0 then θ = nπ + (- 1)n (π/6).

The solutions in the interval [0, 2π) are 0, π, π/6 and 5π/6.

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