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

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

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(2cos(θ)+1)(cos(θ)-2)=0

(2 cos (θ) + 1) (cos (θ) - 2) = 0

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

Solution steps

(2 cos (θ) + 1) (cos (θ) - 2) = 0

Solving each part separately

2 cos (θ) + 1 = 0 or cos (θ) - 2 = 0

2 cos (θ) + 1 = 0 : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) - 2 = 0 :

No Solution

Combine all the solutions

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

4 cos (2 x) - 1 = 1

sin (2 x) = cot (x)

cos (θ) = 3

cos (θ) = 4

7 cos (x) + 9 = - 2 cos (x)

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