What is the general solution for 2sin^2(t)-3sin(t)+1=0 ?

The general solution for 2sin^2(t)-3sin(t)+1=0 is t= pi/2+2pin,t= pi/6+2pin,t=(5pi)/6+2pin

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2sin^2(t)-3sin(t)+1=0

2 sin^{2} (t) - 3 sin (t) + 1 = 0

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

Solution steps

2 sin^{2} (t) - 3 sin (t) + 1 = 0

Solve by substitution

sin (t) = 1, sin (t) = \frac{1}{2}

sin (t) = 1 : t = \frac{π}{2} + 2 πn

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

Combine all the solutions

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

5 sin (x) = 0

5 sin (x) = 4

5 sin (x) = 3

sin (x) (2 cos (x) - 3) = 0

cos^{2} (x) + 5 cos (x) = cos (x) - 3

What is the general solution for 2sin^2(t)-3sin(t)+1=0 ?
The general solution for 2sin^2(t)-3sin(t)+1=0 is t= pi/2+2pin,t= pi/6+2pin,t=(5pi)/6+2pin