Calculate the possible values of Cosθ if 4cos 2θ+ 2 cosθ+3 = 0
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cos2θ = 2*(cos θ)^2 - 1
4*(2*(cos θ)^2 - 1 ) + 2 cosθ + 3 = 0
8(cos θ)^2 - 4 + 2 cosθ + 3 = 0
8(cos θ)^2 + 2 cosθ - 1 = 0
8(cos θ)^2 + 4 cosθ - 2 cosθ - 1 = 0
4 cosθ [ 2cos θ + 1] - [ 2cos θ + 1]= 0
[4 cosθ -1] [ 2cos θ + 1] = 0
Therefore,
[4 cosθ -1] =0 Or [ 2cos θ + 1] = 0
cosθ=1/4 Or cosθ= -1/2
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4*(2*(cos θ)^2 - 1 ) + 2 cosθ + 3 = 0
8(cos θ)^2 - 4 + 2 cosθ + 3 = 0
8(cos θ)^2 + 2 cosθ - 1 = 0
8(cos θ)^2 + 4 cosθ - 2 cosθ - 1 = 0
4 cosθ [ 2cos θ + 1] - [ 2cos θ + 1]= 0
[4 cosθ -1] [ 2cos θ + 1] = 0
Therefore,
[4 cosθ -1] =0 Or [ 2cos θ + 1] = 0
cosθ=1/4 Or cosθ= -1/2
Please mark my answer as brainliest
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