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

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

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2cos^2(θ)-5cos(θ)-3=0

2 cos^{2} (θ) - 5 cos (θ) - 3 = 0

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

Solution steps

2 cos^{2} (θ) - 5 cos (θ) - 3 = 0

Solve by substitution

cos (θ) = 3, cos (θ) = - \frac{1}{2}

cos (θ) = 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

15 sin (x) + 6 cos (x) - 3 = 0

1 + cot^{2} (x) = 8

\frac{1}{1 - sin ( x )} + \frac{1}{1 + sin ( x )} = 2 csc^{2} (x)

tan (x) - 1 - 2 tan^{2} (x) = 0

cos (θ) = - \frac{2}{5}, tan (2 θ)

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