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

The general solution for 2cos^2(θ)+2cos(θ)-1=0 is θ=1.19606…+2pin,θ=2pi-1.19606…+2pin

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2cos^2(θ)+2cos(θ)-1=0

2 cos^{2} (θ) + 2 cos (θ) - 1 = 0

θ = 1.19606 \dots + 2 πn, θ = 2 π - 1.19606 \dots + 2 πn

Solution steps

2 cos^{2} (θ) + 2 cos (θ) - 1 = 0

Solve by substitution

cos (θ) = \frac{- 1 + 3}{2}, cos (θ) = - \frac{1 + 3}{2}

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

cos (θ) = - \frac{1 + 3}{2} :

No Solution

Combine all the solutions

θ = arccos (\frac{- 1 + 3}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 3}{2}) + 2 πn

Show solutions in decimal form

θ = 1.19606 \dots + 2 πn, θ = 2 π - 1.19606 \dots + 2 πn

- 2 sin (- 3 x) - 1 = 0

tan (x) + sin (x) tan (x) = 0, - π \leq x < π

4 sin^{2} (x) - sin (x) - 2 = 0

2 cos (x) sin (x) + 1 cos (x) = 0

sin^{2} (x) = - 2 cos^{2} (x)

What is the general solution for 2cos^2(θ)+2cos(θ)-1=0 ?
The general solution for 2cos^2(θ)+2cos(θ)-1=0 is θ=1.19606…+2pin,θ=2pi-1.19606…+2pin