What is the general solution for 2cos^2(θ)+cos(θ)-1=0 ?

The general solution for 2cos^2(θ)+cos(θ)-1=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin,θ=pi+2pin

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2cos^2(θ)+cos(θ)-1=0

2 cos^{2} (θ) + cos (θ) - 1 = 0

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = π + 2 πn

Solution steps

2 cos^{2} (θ) + cos (θ) - 1 = 0

Solve by substitution

cos (θ) = \frac{1}{2}, cos (θ) = - 1

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = - 1 : θ = π + 2 πn

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = π + 2 πn

2 sin^{2} (θ) + cos (θ) = 1

2 cos (2 x) - 2 cos (x) = 0

sec (x) + 2 = 0

cos (x) - sin (2 x) = 0

csc (x) = - 1

What is the general solution for 2cos^2(θ)+cos(θ)-1=0 ?
The general solution for 2cos^2(θ)+cos(θ)-1=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin,θ=pi+2pin