What is the general solution for 9cos(2θ)=8 ?

The general solution for 9cos(2θ)=8 is θ=(0.47588…)/2+pin,θ=pi-(0.47588…)/2+pin

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9cos(2θ)=8

9 cos (2 θ) = 8

θ = \frac{0.47588 \dots}{2} + πn, θ = π - \frac{0.47588 \dots}{2} + πn

Solution steps

9 cos (2 θ) = 8

Divide both sides by

9

cos (2 θ) = \frac{8}{9}

Apply trig inverse properties

2 θ = arccos (\frac{8}{9}) + 2 πn, 2 θ = 2 π - arccos (\frac{8}{9}) + 2 πn

Solve

2 θ = arccos (\frac{8}{9}) + 2 πn : θ = \frac{arccos ( \frac{8}{9} )}{2} + πn

Solve

2 θ = 2 π - arccos (\frac{8}{9}) + 2 πn : θ = π - \frac{arccos ( \frac{8}{9} )}{2} + πn

θ = \frac{arccos ( \frac{8}{9} )}{2} + πn, θ = π - \frac{arccos ( \frac{8}{9} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.47588 \dots}{2} + πn, θ = π - \frac{0.47588 \dots}{2} + πn

cos (2 θ) + 6 sin^{2} (θ) = 4, 0 \leq θ < 2 π

tan (x) = \frac{6}{9}

3 csc (θ) - 2 = 0

2 tan (2 x) + 2 = 0

2 sin^{2} (x) - 3 = 0

What is the general solution for 9cos(2θ)=8 ?
The general solution for 9cos(2θ)=8 is θ=(0.47588…)/2+pin,θ=pi-(0.47588…)/2+pin