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

The general solution for 2sin^2(θ)-sin(θ)-2=0 is θ=-0.89590…+2pin,θ=pi+0.89590…+2pin

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2sin^2(θ)-sin(θ)-2=0

2 sin^{2} (θ) - sin (θ) - 2 = 0

θ = - 0.89590 \dots + 2 πn, θ = π + 0.89590 \dots + 2 πn

Solution steps

2 sin^{2} (θ) - sin (θ) - 2 = 0

Solve by substitution

sin (θ) = \frac{1 + 17}{4}, sin (θ) = \frac{1 - 17}{4}

sin (θ) = \frac{1 + 17}{4} :

No Solution

sin (θ) = \frac{1 - 17}{4} : θ = arcsin (\frac{1 - 17}{4}) + 2 πn, θ = π + arcsin (- \frac{1 - 17}{4}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{1 - 17}{4}) + 2 πn, θ = π + arcsin (- \frac{1 - 17}{4}) + 2 πn

Show solutions in decimal form

θ = - 0.89590 \dots + 2 πn, θ = π + 0.89590 \dots + 2 πn

3 cot (x) + 2 = 5

sec^{2} (x) - tan (x) - 1 = 0

2 cos (2 x + \frac{π}{6}) + 3 = 2, 0 < x < 2 π

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

(cot (θ) + 3) (csc (θ) + 2) = 0

What is the general solution for 2sin^2(θ)-sin(θ)-2=0 ?
The general solution for 2sin^2(θ)-sin(θ)-2=0 is θ=-0.89590…+2pin,θ=pi+0.89590…+2pin