What is the general solution for 3cos^2(θ)+6cos(θ)-4=0 ?

The general solution for 3cos^2(θ)+6cos(θ)-4=0 is θ=1.01511…+2pin,θ=2pi-1.01511…+2pin

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3cos^2(θ)+6cos(θ)-4=0

3 cos^{2} (θ) + 6 cos (θ) - 4 = 0

θ = 1.01511 \dots + 2 πn, θ = 2 π - 1.01511 \dots + 2 πn

Solution steps

3 cos^{2} (θ) + 6 cos (θ) - 4 = 0

Solve by substitution

cos (θ) = \frac{- 3 + 21}{3}, cos (θ) = - \frac{3 + 21}{3}

cos (θ) = \frac{- 3 + 21}{3} : θ = arccos (\frac{- 3 + 21}{3}) + 2 πn, θ = 2 π - arccos (\frac{- 3 + 21}{3}) + 2 πn

cos (θ) = - \frac{3 + 21}{3} :

No Solution

Combine all the solutions

θ = arccos (\frac{- 3 + 21}{3}) + 2 πn, θ = 2 π - arccos (\frac{- 3 + 21}{3}) + 2 πn

Show solutions in decimal form

θ = 1.01511 \dots + 2 πn, θ = 2 π - 1.01511 \dots + 2 πn

arccos (x) = 1

cos (x) = \frac{3}{4}

sin (x) = sin (\frac{x}{2})

sec (x) - 2 = - tan (x) - 3

- 9.81 + 5 (32.17) (\frac{12}{2}) sin (θ) = 0

What is the general solution for 3cos^2(θ)+6cos(θ)-4=0 ?
The general solution for 3cos^2(θ)+6cos(θ)-4=0 is θ=1.01511…+2pin,θ=2pi-1.01511…+2pin