What is the general solution for 5cos(2θ)=15cos(θ)-10 ?

The general solution for 5cos(2θ)=15cos(θ)-10 is θ=2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin

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5cos(2θ)=15cos(θ)-10

5 cos (2 θ) = 15 cos (θ) - 10

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

Solution steps

5 cos (2 θ) = 15 cos (θ) - 10

Subtract

15 cos (θ) - 10

from both sides

5 cos (2 θ) - 15 cos (θ) + 10 = 0

Rewrite using trig identities

5 + 10 cos^{2} (θ) - 15 cos (θ) = 0

Solve by substitution

cos (θ) = 1, cos (θ) = \frac{1}{2}

cos (θ) = 1 : θ = 2 πn

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

Combine all the solutions

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

cos^{2} (x) = 1 + sin^{2} (x)

2 cos^{2} (x) + cos (x) = 0, 0 \leq x \leq 2 π

sin^{2} (x) - 2 cos (x) = 2

2 cos (θ) - sin (θ) = 0

cos (x + \frac{π}{6}) = sin (x)

What is the general solution for 5cos(2θ)=15cos(θ)-10 ?
The general solution for 5cos(2θ)=15cos(θ)-10 is θ=2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin