What is the general solution for -5cos^2(θ)-4cos(θ)+4=0 ?

The general solution for -5cos^2(θ)-4cos(θ)+4=0 is θ=0.95231…+2pin,θ=2pi-0.95231…+2pin

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-5cos^2(θ)-4cos(θ)+4=0

- 5 cos^{2} (θ) - 4 cos (θ) + 4 = 0

θ = 0.95231 \dots + 2 πn, θ = 2 π - 0.95231 \dots + 2 πn

Solution steps

- 5 cos^{2} (θ) - 4 cos (θ) + 4 = 0

Solve by substitution

cos (θ) = - \frac{2 ( 1 + 6 )}{5}, cos (θ) = \frac{2 ( 6 - 1 )}{5}

cos (θ) = - \frac{2 ( 1 + 6 )}{5} :

No Solution

cos (θ) = \frac{2 ( 6 - 1 )}{5} : θ = arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn

Combine all the solutions

θ = arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn, θ = 2 π - arccos (\frac{2 ( 6 - 1 )}{5}) + 2 πn

Show solutions in decimal form

θ = 0.95231 \dots + 2 πn, θ = 2 π - 0.95231 \dots + 2 πn

cos (x) = 0.159

sin (2 x) = 1 - cos (2 x)

- 1 = cos (θ)

sin (θ) sec (θ) = tan^{2} (θ)

sin (3 x) - 2 sin (x) = 0

What is the general solution for -5cos^2(θ)-4cos(θ)+4=0 ?
The general solution for -5cos^2(θ)-4cos(θ)+4=0 is θ=0.95231…+2pin,θ=2pi-0.95231…+2pin