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

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

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2sin^2(θ)-sin(θ)-1=0

2 sin^{2} (θ) - sin (θ) - 1 = 0

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

Solution steps

2 sin^{2} (θ) - sin (θ) - 1 = 0

Solve by substitution

sin (θ) = 1, sin (θ) = - \frac{1}{2}

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

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

Combine all the solutions

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

tan (7 5^{\circ})

cos (21 0^{\circ})

cos^{2} (x) = 2 - 2 sin (x)

cos (2 π)

cos (β) = 0.8

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