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

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

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

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

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn, t = \frac{3 π}{2} + 2 πn

Solution steps

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

Solve by substitution

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

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

sin (t) = - 1 : t = \frac{3 π}{2} + 2 πn

Combine all the solutions

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn, t = \frac{3 π}{2} + 2 πn

sin (θ) + sin (3 θ) = 0

4 sin^{2} (x) - 1 = 0, 0 \leq x < 2 π

cos^{2} (x) - 4 sin^{2} (x) = 2

sin (x) = 0.53

2 cos (2 θ) = 2

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