What is the general solution for 2cos(2θ)+1=-12cos(θ)-6 ?

The general solution for 2cos(2θ)+1=-12cos(θ)-6 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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2cos(2θ)+1=-12cos(θ)-6

2 cos (2 θ) + 1 = - 12 cos (θ) - 6

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

Solution steps

2 cos (2 θ) + 1 = - 12 cos (θ) - 6

Subtract

- 12 cos (θ) - 6

from both sides

2 cos (2 θ) + 12 cos (θ) + 7 = 0

Rewrite using trig identities

5 + 12 cos (θ) + 4 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = - \frac{1}{2}, cos (θ) = - \frac{5}{2}

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

cos (θ) = - \frac{5}{2} :

No Solution

Combine all the solutions

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

- 3 cos (2 θ) - 9 cos (θ) - 2 = 2 cos (θ) - 6

23 sin (x^{2}) = 3

cos (4 α) - cos (2 α) = sin (3 α)

3 sec (2 x) + 2 = 0, 0 \leq x < 2 π

sin (x) + cos (x) = 1, 0 \leq x \leq 2 π

What is the general solution for 2cos(2θ)+1=-12cos(θ)-6 ?
The general solution for 2cos(2θ)+1=-12cos(θ)-6 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin