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

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

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-2cos^2(x)-cos(x)+1=0

- 2 cos^{2} (x) - cos (x) + 1 = 0

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

Solution steps

- 2 cos^{2} (x) - cos (x) + 1 = 0

Solve by substitution

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

cos (x) = - 1 : x = π + 2 πn

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

Combine all the solutions

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

sin (x) = \frac{9}{14}

4 tan^{2} (x) + 5 tan (x) - 6 = 0

sin (x) = \frac{1}{2}, 0 \leq x < 2 π

cos (t) + cos (2 t) = - 1

sin (θ) = \frac{3}{4}

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