What is the general solution for 2sin^2(θ)-5sin(θ)-3=0 ?

The general solution for 2sin^2(θ)-5sin(θ)-3=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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2sin^2(θ)-5sin(θ)-3=0

2 sin^{2} (θ) - 5 sin (θ) - 3 = 0

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

Solution steps

2 sin^{2} (θ) - 5 sin (θ) - 3 = 0

Solve by substitution

sin (θ) = 3, sin (θ) = - \frac{1}{2}

sin (θ) = 3 :

No Solution

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

4 cos^{2} (x) - 4 cos (x) - 1 = 0, 0 \leq x \leq 2 π

10 = 6.7 + 13.75 tan (θ) - 4.718 (1 + tan^{2} (θ))

2 cos (3 θ) = - 3

cos (θ) = - cos (θ)

cot (x) cos (x) = cos (x)

What is the general solution for 2sin^2(θ)-5sin(θ)-3=0 ?
The general solution for 2sin^2(θ)-5sin(θ)-3=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin