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

The general solution for 2cos^2(θ)+sin(θ)-1=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ= pi/2+2pin

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2cos^2(θ)+sin(θ)-1=0

2 cos^{2} (θ) + sin (θ) - 1 = 0

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{π}{2} + 2 πn

Solution steps

2 cos^{2} (θ) + sin (θ) - 1 = 0

Rewrite using trig identities

1 + sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - \frac{1}{2}, sin (θ) = 1

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{π}{2} + 2 πn

2 sin (2 x) sin (x) - 3 cos (x) = 0

sin (θ) = \frac{13}{7}

sin^{2} (θ) - cos^{2} (θ) = 1

2 sin (3 θ) = 2

cot^{2} (x) + 2 cot (x) + 1 = 0

What is the general solution for 2cos^2(θ)+sin(θ)-1=0 ?
The general solution for 2cos^2(θ)+sin(θ)-1=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ= pi/2+2pin