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

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

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2cos^2(θ)-5cos(θ)+2=0

2 cos^{2} (θ) - 5 cos (θ) + 2 = 0

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

Solution steps

2 cos^{2} (θ) - 5 cos (θ) + 2 = 0

Solve by substitution

cos (θ) = 2, cos (θ) = \frac{1}{2}

cos (θ) = 2 :

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

tan^{2} (x) - 3 tan (x) - 4 = 0

2 cos (x) + 4 = 5

cos (2 x - \frac{π}{2}) = - 1

cot (x) = - \frac{1}{3}

4 cos^{2} (x) + 2 cos (x) - 2 = 0

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