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

The general solution for 9cos(2θ)=9cos(θ) is θ=2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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9cos(2θ)=9cos(θ)

9 cos (2 θ) = 9 cos (θ)

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

Solution steps

9 cos (2 θ) = 9 cos (θ)

Subtract

9 cos (θ)

from both sides

9 cos (2 θ) - 9 cos (θ) = 0

Rewrite using trig identities

(- 1 + 2 cos^{2} (θ)) \cdot 9 - 9 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{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Combine all the solutions

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

6 sin (x) + 4 cos (x) = 0

cos (t) + sin (t) = 0

cos (C) = 0.87

4 cot (x) + 3 = - 1

cos (\frac{θ}{2}) - cos (θ) = 1

What is the general solution for 9cos(2θ)=9cos(θ) ?
The general solution for 9cos(2θ)=9cos(θ) is θ=2pin,θ=(2pi)/3+2pin,θ=(4pi)/3+2pin