What is the general solution for 2+8cos(θ)=-1+2cos(θ) ?

The general solution for 2+8cos(θ)=-1+2cos(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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2+8cos(θ)=-1+2cos(θ)

2 + 8 cos (θ) = - 1 + 2 cos (θ)

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

Solution steps

2 + 8 cos (θ) = - 1 + 2 cos (θ)

Solve by substitution

cos (θ) = - \frac{1}{2}

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

Combine all the solutions

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

sin^{2} (x) + cos^{2} (x) + cos (x) = 0

2.5 = 10 sin (x)

4.797896 cos (0.568 x - 2.595) = 0

2 cos (x) - 1 = 2 cos (2 x)

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

What is the general solution for 2+8cos(θ)=-1+2cos(θ) ?
The general solution for 2+8cos(θ)=-1+2cos(θ) is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin