What is the general solution for 2sin^2(θ)-7sin(θ)+5=0 ?

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

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2sin^2(θ)-7sin(θ)+5=0

2 sin^{2} (θ) - 7 sin (θ) + 5 = 0

θ = \frac{π}{2} + 2 πn

Solution steps

2 sin^{2} (θ) - 7 sin (θ) + 5 = 0

Solve by substitution

sin (θ) = \frac{5}{2}, sin (θ) = 1

sin (θ) = \frac{5}{2} :

No Solution

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn

x, 4 cos^{2} (x) - 3 = 0

tan (x) = \frac{2}{4}

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

4 sin^{2} (x) + 4 sin (x) = - 1

sin (θ) = \frac{8}{17}, cos (θ)

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