What is the general solution for 3cos(2θ)-5cos(θ)-1=0 ?

The general solution for 3cos(2θ)-5cos(θ)-1=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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3cos(2θ)-5cos(θ)-1=0

3 cos (2 θ) - 5 cos (θ) - 1 = 0

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

Solution steps

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

3 cos (b) + 3 = 0

csc (x) = \frac{21}{5}

3 cot (2 x) + 3 = 0

\frac{1}{cos ( θ )} = 2

θ, r = sin (2 θ)

What is the general solution for 3cos(2θ)-5cos(θ)-1=0 ?
The general solution for 3cos(2θ)-5cos(θ)-1=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin