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2cos^4x-3cos^2x=-1

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asked Jul 18, 2014 in TRIGONOMETRY by anonymous

1 Answer

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2cos^4x-3cos^2x=-1

The trigonometric equation is 2cos4 (x) - 3cos2 (x) + 1 = 0.

Let, cos2 (x) = t.

Then, the equation is 2t2 - 3t + 1 = 0.

2t2 - 2t - t + 1 = 0

2t(t - 1) - 1(t - 1) = 0

(t - 1)(2t - 1) = 0

⇒t - 1 = 0 and 2t - 1 = 0

t =  1 and t = 1/2

cos2 (x) = 1 and cos2 (x) = 1/2.

cos (x) = ±1 and cos2 (x) =± 1/√2.

cos (x) = 1

cos (x) = cos(0)

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

x = 2nπ ± 0

x = 2nπ ; n = 0,1,2..........

cos (x) =-1

cos (x) = cos(π)

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

x = 2nπ ± π

x =(2n±1)π ; n = 0,1,2..........

cos (x) =1/√2.

cos (x) = cos(π/4)

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

x = 2nπ ± (π/4)

   =(2n ± 1/4)π;  n = 0,1,2..........

cos (x) =-1/√2.

cos (x) = cos(3π/4)

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

x = 2nπ ± (3π/4)

   =(2n ± 3/4)π ; n = 0,1,2..........

answered Jul 18, 2014 by bradely Mentor

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