What is the general solution for 3cos^2(θ)+4cos(θ)-4=0 ?

The general solution for 3cos^2(θ)+4cos(θ)-4=0 is θ=0.84106…+2pin,θ=2pi-0.84106…+2pin

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3cos^2(θ)+4cos(θ)-4=0

3 cos^{2} (θ) + 4 cos (θ) - 4 = 0

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

Solution steps

3 cos^{2} (θ) + 4 cos (θ) - 4 = 0

Solve by substitution

cos (θ) = \frac{2}{3}, cos (θ) = - 2

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

cos (θ) = - 2 :

No Solution

Combine all the solutions

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

Show solutions in decimal form

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

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

2 sin (x) cos (x) = 3 sin (x)

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

sin (5 x + 50) = cos (4 x - 23)

tan (x) = 1.2

What is the general solution for 3cos^2(θ)+4cos(θ)-4=0 ?
The general solution for 3cos^2(θ)+4cos(θ)-4=0 is θ=0.84106…+2pin,θ=2pi-0.84106…+2pin