What is the general solution for 9cos(2θ)=27cos(θ)-18 ?

The general solution for 9cos(2θ)=27cos(θ)-18 is θ=2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin

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9cos(2θ)=27cos(θ)-18

9 cos (2 θ) = 27 cos (θ) - 18

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

Solution steps

9 cos (2 θ) = 27 cos (θ) - 18

Subtract

27 cos (θ) - 18

from both sides

9 cos (2 θ) - 27 cos (θ) + 18 = 0

Rewrite using trig identities

9 + 18 cos^{2} (θ) - 27 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

4 csc (x) = 3 csc (x) cot (x) + 3 csc (x)

4 cos^{2} (x) + 4 cos (x) = 1, 0^{\circ} \leq x < 36 0^{\circ}

8 cos^{2} (x) = 4

4 cos (3 x) = 0

sin^{2} (x) = sin^{2} (\frac{x}{2})

What is the general solution for 9cos(2θ)=27cos(θ)-18 ?
The general solution for 9cos(2θ)=27cos(θ)-18 is θ=2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin