What is the general solution for 2cos^2(θ)-3cos(θ)-2=0 ?

The general solution for 2cos^2(θ)-3cos(θ)-2=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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2cos^2(θ)-3cos(θ)-2=0

2 cos^{2} (θ) - 3 cos (θ) - 2 = 0

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Solution steps

2 cos^{2} (θ) - 3 cos (θ) - 2 = 0

Solve by substitution

cos (θ) = 2, cos (θ) = - \frac{1}{2}

cos (θ) = 2 :

No Solution

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Combine all the solutions

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

- 7 sin^{2} (θ) + 4 sin (θ) = - 3

3 = 3 cos (2 x) + sin (2 x)

csc (θ) = - \frac{2 3}{3}

2 sin (\frac{x}{3}) + 3 = 0

tan (θ) = \frac{15}{8}

What is the general solution for 2cos^2(θ)-3cos(θ)-2=0 ?
The general solution for 2cos^2(θ)-3cos(θ)-2=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin