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

The general solution for 2(sin(t))^2-sin(t)-1=0 is t= pi/2+2pin,t=(7pi)/6+2pin,t=(11pi)/6+2pin

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

2 (sin (t))^{2} - 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 (t))^{2} - 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) = 0.625

0 = 2 cos (t)

sin (z) \cdot cos (z) = 1

0 = 2 cos (x)

\frac{cos ( x )}{cos ( x )} = \frac{cos ( x )}{1 + sin ( x )}

What is the general solution for 2(sin(t))^2-sin(t)-1=0 ?
The general solution for 2(sin(t))^2-sin(t)-1=0 is t= pi/2+2pin,t=(7pi)/6+2pin,t=(11pi)/6+2pin