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

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

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2sin^2(θ)-cos(θ)-1=0

2 sin^{2} (θ) - cos (θ) - 1 = 0

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

Solution steps

2 sin^{2} (θ) - cos (θ) - 1 = 0

Rewrite using trig identities

1 - cos (θ) - 2 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = - 1, cos (θ) = \frac{1}{2}

cos (θ) = - 1 : θ = π + 2 πn

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

Combine all the solutions

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

cos (x) + 1 = 2 sin^{2} (x)

8 sec^{2} (\frac{x}{2}) = 4

6 cos^{2} (x) + cos (x) = 1

4 sin (x) + 3 cos (x) = 0

y, x = 2 sin (y)

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