What is the general solution for 3cos(2θ)-5cos(θ)=1 ?

The general solution for 3cos(2θ)-5cos(θ)=1 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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3cos(2θ)-5cos(θ)=1

3 cos (2 θ) - 5 cos (θ) = 1

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

Solution steps

3 cos (2 θ) - 5 cos (θ) = 1

Subtract

1

from both sides

3 cos (2 θ) - 5 cos (θ) - 1 = 0

Rewrite using trig identities

- 4 - 5 cos (θ) + 6 cos^{2} (θ) = 0

Solve by substitution

cos (θ) = \frac{4}{3}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{4}{3} :

No Solution

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) = 4 cos (2 x)

- 9 cos^{2} (θ) + 1 = sin (θ)

1 - sin (θ) = sin (θ)

cot (\frac{θ}{3} + \frac{3 π}{4}) = 3

cos (x) = \frac{tan ( x ) + cot ( x )}{csc ( x )}

What is the general solution for 3cos(2θ)-5cos(θ)=1 ?
The general solution for 3cos(2θ)-5cos(θ)=1 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin