Solve 3 cosx + 3 = 2 sin^2 x on the interval [0, 2pi)

Calculus
Tutor: None Selected Time limit: 1 Day

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Jan 28th, 2015

3 cos x + 3 = 2 sin^2 x
3 cos x +3 = 2( 1- cos^2 x)
3 cos x + 3 = 2 - 2 cos^2x
2cos^2 x+3 cos x +1 = 0
2 cos^2 + 2 cos x + cos x + 1 = 0

2cos x ( cos x +1)+1 (cos x+1) = 0
(2cos x +1)(cos x + 1) =0

therefore,
2cos x + 1 = 0 and cos x+1 = 0

so, 2cos x=-1 and cos x =-1
cos x = -1/2 and cos x =-1
hence,
x = cos^-1 (-1/2) and x = cos^-1 (-1)
x = 2pi/3 and x = pi

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Jan 28th, 2015

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