What is the general solution for 2cos^2(θ)-5cos(θ)+1=0 ?

The general solution for 2cos^2(θ)-5cos(θ)+1=0 is θ=1.34977…+2pin,θ=2pi-1.34977…+2pin

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2cos^2(θ)-5cos(θ)+1=0

2 cos^{2} (θ) - 5 cos (θ) + 1 = 0

θ = 1.34977 \dots + 2 πn, θ = 2 π - 1.34977 \dots + 2 πn

Solution steps

2 cos^{2} (θ) - 5 cos (θ) + 1 = 0

Solve by substitution

cos (θ) = \frac{5 + 17}{4}, cos (θ) = \frac{5 - 17}{4}

cos (θ) = \frac{5 + 17}{4} :

No Solution

cos (θ) = \frac{5 - 17}{4} : θ = arccos (\frac{5 - 17}{4}) + 2 πn, θ = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

Combine all the solutions

θ = arccos (\frac{5 - 17}{4}) + 2 πn, θ = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

Show solutions in decimal form

θ = 1.34977 \dots + 2 πn, θ = 2 π - 1.34977 \dots + 2 πn

- 2 cos^{2} (x) + sin (x) + 1 = 0

tan (x) + sec (x) = 1, 0 \leq x \leq 2 π

8 sin (θ) cos (θ) = 3

tan (x) = \frac{7}{9}

cos (8 x) = - 1

What is the general solution for 2cos^2(θ)-5cos(θ)+1=0 ?
The general solution for 2cos^2(θ)-5cos(θ)+1=0 is θ=1.34977…+2pin,θ=2pi-1.34977…+2pin