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

The general solution for 2cos^2(t)-cos(t)-1=0 is t=2pin,t=(2pi)/3+2pin,t=(4pi)/3+2pin

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

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

t = 2 πn, t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

Solution steps

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

Solve by substitution

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

cos (t) = 1 : t = 2 πn

cos (t) = - \frac{1}{2} : t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

Combine all the solutions

t = 2 πn, t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

sin (x) = - \frac{8}{17}

1 - 2 cos (θ) = 0

2 sin (\frac{x}{2}) - 2 = 0

- 3 sin (3 θ) = 0

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

What is the general solution for 2cos^2(t)-cos(t)-1=0 ?
The general solution for 2cos^2(t)-cos(t)-1=0 is t=2pin,t=(2pi)/3+2pin,t=(4pi)/3+2pin