What is the general solution for 2cos(2θ)-4cos(θ)+2=-1 ?

The general solution for 2cos(2θ)-4cos(θ)+2=-1 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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2cos(2θ)-4cos(θ)+2=-1

2 cos (2 θ) - 4 cos (θ) + 2 = - 1

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

Solution steps

2 cos (2 θ) - 4 cos (θ) + 2 = - 1

Subtract

- 1

from both sides

2 cos (2 θ) - 4 cos (θ) + 3 = 0

Rewrite using trig identities

1 - 4 cos (θ) + 4 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = \frac{1}{2}

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

Combine all the solutions

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

cos (a) = \frac{1}{sin ( a )}

sin (a) = cos (a)

tan (a) = 2.0503

4 sin (x) = 3 csc (x) - 4

5 sin (θ) - 3 = 0

What is the general solution for 2cos(2θ)-4cos(θ)+2=-1 ?
The general solution for 2cos(2θ)-4cos(θ)+2=-1 is θ= pi/3+2pin,θ=(5pi)/3+2pin