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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

cos (t) = - 1 : t = π + 2 πn

cos (t) = \frac{1}{2} : t = \frac{π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn

Combine all the solutions

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

4 cos^{2} (θ) = 1, 0 \leq θ < 2 π

cos (\frac{x}{2} + \frac{π}{3}) = \frac{2}{2}

tan (θ) = 8.2

2 sin (x) = 4 sin (x) + 1

- 1 - cos (θ) = - 3 cos (θ)

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