What is the general solution for 5sin^2(θ)-11sin(θ)+2=0 ?

The general solution for 5sin^2(θ)-11sin(θ)+2=0 is θ=0.20135…+2pin,θ=pi-0.20135…+2pin

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5sin^2(θ)-11sin(θ)+2=0

5 sin^{2} (θ) - 11 sin (θ) + 2 = 0

θ = 0.20135 \dots + 2 πn, θ = π - 0.20135 \dots + 2 πn

Solution steps

5 sin^{2} (θ) - 11 sin (θ) + 2 = 0

Solve by substitution

sin (θ) = 2, sin (θ) = \frac{1}{5}

sin (θ) = 2 :

No Solution

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

Combine all the solutions

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

Show solutions in decimal form

θ = 0.20135 \dots + 2 πn, θ = π - 0.20135 \dots + 2 πn

tan^{2} (θ) + 3 sec (θ) = - 3

cos^{2} (x) - sin^{2} (x) = 1 - sin^{2} (x)

sin (2 x) cos (x) + sin (x) = 0

cos (2 x) + cos (x) = 1 + sin (2 x) - sin (x)

- sin (θ) + 2 sin^{2} (θ) = 0

What is the general solution for 5sin^2(θ)-11sin(θ)+2=0 ?
The general solution for 5sin^2(θ)-11sin(θ)+2=0 is θ=0.20135…+2pin,θ=pi-0.20135…+2pin