What is the general solution for cos(2θ)= 20/29 ?

The general solution for cos(2θ)= 20/29 is θ=(0.80978…)/2+pin,θ=pi-(0.80978…)/2+pin

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cos(2θ)= 20/29

cos (2 θ) = \frac{20}{29}

θ = \frac{0.80978 \dots}{2} + πn, θ = π - \frac{0.80978 \dots}{2} + πn

Degrees

θ = 23.19859 \dots^{\circ} + 18 0^{\circ} n, θ = 156.80140 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 θ) = \frac{20}{29}

Apply trig inverse properties

cos (2 θ) = \frac{20}{29}

General solutions for

cos (2 θ) = \frac{20}{29}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

2 θ = arccos (\frac{20}{29}) + 2 πn, 2 θ = 2 π - arccos (\frac{20}{29}) + 2 πn

2 θ = arccos (\frac{20}{29}) + 2 πn, 2 θ = 2 π - arccos (\frac{20}{29}) + 2 πn

Solve

2 θ = arccos (\frac{20}{29}) + 2 πn : θ = \frac{arccos ( \frac{20}{29} )}{2} + πn

2 θ = arccos (\frac{20}{29}) + 2 πn

Divide both sides by

2

2 θ = arccos (\frac{20}{29}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arccos ( \frac{20}{29} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{arccos ( \frac{20}{29} )}{2} + πn

θ = \frac{arccos ( \frac{20}{29} )}{2} + πn

Solve

2 θ = 2 π - arccos (\frac{20}{29}) + 2 πn : θ = π - \frac{arccos ( \frac{20}{29} )}{2} + πn

2 θ = 2 π - arccos (\frac{20}{29}) + 2 πn

Divide both sides by

2

2 θ = 2 π - arccos (\frac{20}{29}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{20}{29} )}{2} + \frac{2 πn}{2}

Simplify

θ = π - \frac{arccos ( \frac{20}{29} )}{2} + πn

θ = π - \frac{arccos ( \frac{20}{29} )}{2} + πn

θ = \frac{arccos ( \frac{20}{29} )}{2} + πn, θ = π - \frac{arccos ( \frac{20}{29} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.80978 \dots}{2} + πn, θ = π - \frac{0.80978 \dots}{2} + πn

- 6 sin (3 x) = 0

sin (x) = \frac{2}{4}

2 tan (2 x) - 6 = 0

8 sin (θ) + 15 cos (θ) = 18

\frac{30}{sin ( A )} = \frac{34.4}{sin ( 6 2 ^{\circ} )}

What is the general solution for cos(2θ)= 20/29 ?
The general solution for cos(2θ)= 20/29 is θ=(0.80978…)/2+pin,θ=pi-(0.80978…)/2+pin