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/2+2pin,t=(7pi)/6+2pin,t=(11pi)/6+2pin

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

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

t = \frac{π}{2} + 2 πn, t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

Solution steps

2 sin^{2} (t) - 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{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

Combine all the solutions

t = \frac{π}{2} + 2 πn, t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

cos (x) = - \frac{1}{5 2}

cot (x) tan (2 x) = 3

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

sin (3 x) = sin (5 x)

2 sin^{2} (3 θ) + sin (3 θ) - 1 = 0

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/2+2pin,t=(7pi)/6+2pin,t=(11pi)/6+2pin