What is the general solution for 8cos^2(θ)+13=14cos(θ)+8 ?

The general solution for 8cos^2(θ)+13=14cos(θ)+8 is θ= pi/3+2pin,θ=(5pi)/3+2pin

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8cos^2(θ)+13=14cos(θ)+8

8 cos^{2} (θ) + 13 = 14 cos (θ) + 8

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

Solution steps

8 cos^{2} (θ) + 13 = 14 cos (θ) + 8

Solve by substitution

cos (θ) = \frac{5}{4}, cos (θ) = \frac{1}{2}

cos (θ) = \frac{5}{4} :

No Solution

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

Combine all the solutions

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

cos (3 x) = - \frac{1}{2}, 0 \leq x < 2 π

cos (5 t) = 0

tan (4 x) = \frac{3}{3}

cos (v) = \frac{1}{2}

sin (θ) (2 cos (θ) + 1) = 0

What is the general solution for 8cos^2(θ)+13=14cos(θ)+8 ?
The general solution for 8cos^2(θ)+13=14cos(θ)+8 is θ= pi/3+2pin,θ=(5pi)/3+2pin