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=(3pi)/2+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

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2cos^2(θ)-sin(θ)-1=0

2 cos^{2} (θ) - sin (θ) - 1 = 0

θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 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) = - 1, sin (θ) = \frac{1}{2}

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

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (t) = - 1

sin (3 x) = 1, 0 \leq x \leq 2 π

x, z = arcsin (\frac{x}{x ^{2} + y ^{2}})

cos (θ) (2 cos (θ) + 1) = 0

2 cot (x) - 2 = 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=(3pi)/2+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin