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

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

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2sin^2(θ)-5sin(θ)+2=0

2 sin^{2} (θ) - 5 sin (θ) + 2 = 0

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

2 sin^{2} (θ) - 5 sin (θ) + 2 = 0

Solve by substitution

sin (θ) = 2, sin (θ) = \frac{1}{2}

sin (θ) = 2 :

No Solution

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

tan (2 x) + 1 = 0

sin (x) - 2 = - sin (x)

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

8 sec (x) csc (x) = 16 csc (x)

sin (2 t) = 0

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