What is the general solution for 3cos^2(θ)+7cos(θ)-8=0 ?

The general solution for 3cos^2(θ)+7cos(θ)-8=0 is θ=0.57302…+2pin,θ=2pi-0.57302…+2pin

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3cos^2(θ)+7cos(θ)-8=0

3 cos^{2} (θ) + 7 cos (θ) - 8 = 0

θ = 0.57302 \dots + 2 πn, θ = 2 π - 0.57302 \dots + 2 πn

Solution steps

3 cos^{2} (θ) + 7 cos (θ) - 8 = 0

Solve by substitution

cos (θ) = \frac{- 7 + 145}{6}, cos (θ) = \frac{- 7 - 145}{6}

cos (θ) = \frac{- 7 + 145}{6} : θ = arccos (\frac{- 7 + 145}{6}) + 2 πn, θ = 2 π - arccos (\frac{- 7 + 145}{6}) + 2 πn

cos (θ) = \frac{- 7 - 145}{6} :

No Solution

Combine all the solutions

θ = arccos (\frac{- 7 + 145}{6}) + 2 πn, θ = 2 π - arccos (\frac{- 7 + 145}{6}) + 2 πn

Show solutions in decimal form

θ = 0.57302 \dots + 2 πn, θ = 2 π - 0.57302 \dots + 2 πn

2 cos (2 a) = 2, 0 \leq θ \leq 2 π

t, y^{4} - 81 y = sin (t)

11 sin (B) - 1 = 3 sin (B) - 1

sin (5 x) - sin (3 x) - sin (x) = 0

4 arccos (2 x) = π

What is the general solution for 3cos^2(θ)+7cos(θ)-8=0 ?
The general solution for 3cos^2(θ)+7cos(θ)-8=0 is θ=0.57302…+2pin,θ=2pi-0.57302…+2pin