What is the general solution for cos(2θ)= 3/5 ?

The general solution for cos(2θ)= 3/5 is θ=(0.92729…)/2+pin,θ=pi-(0.92729…)/2+pin

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cos(2θ)= 3/5

cos (2 θ) = \frac{3}{5}

θ = \frac{0.92729 \dots}{2} + πn, θ = π - \frac{0.92729 \dots}{2} + πn

Degrees

θ = 26.56505 \dots^{\circ} + 18 0^{\circ} n, θ = 153.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 θ) = \frac{3}{5}

Apply trig inverse properties

cos (2 θ) = \frac{3}{5}

General solutions for

cos (2 θ) = \frac{3}{5}

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

2 θ = arccos (\frac{3}{5}) + 2 πn, 2 θ = 2 π - arccos (\frac{3}{5}) + 2 πn

2 θ = arccos (\frac{3}{5}) + 2 πn, 2 θ = 2 π - arccos (\frac{3}{5}) + 2 πn

Solve

2 θ = arccos (\frac{3}{5}) + 2 πn : θ = \frac{arccos ( \frac{3}{5} )}{2} + πn

2 θ = arccos (\frac{3}{5}) + 2 πn

Divide both sides by

2

2 θ = arccos (\frac{3}{5}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arccos ( \frac{3}{5} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{arccos ( \frac{3}{5} )}{2} + πn

θ = \frac{arccos ( \frac{3}{5} )}{2} + πn

Solve

2 θ = 2 π - arccos (\frac{3}{5}) + 2 πn : θ = π - \frac{arccos ( \frac{3}{5} )}{2} + πn

2 θ = 2 π - arccos (\frac{3}{5}) + 2 πn

Divide both sides by

2

2 θ = 2 π - arccos (\frac{3}{5}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{3}{5} )}{2} + \frac{2 πn}{2}

Simplify

θ = π - \frac{arccos ( \frac{3}{5} )}{2} + πn

θ = π - \frac{arccos ( \frac{3}{5} )}{2} + πn

θ = \frac{arccos ( \frac{3}{5} )}{2} + πn, θ = π - \frac{arccos ( \frac{3}{5} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.92729 \dots}{2} + πn, θ = π - \frac{0.92729 \dots}{2} + πn

cos (\frac{x}{3}) = \frac{1}{2}

3 cos (2 x) - 5 cos (x) = 1

3 tan^{2} (x) - sec (x) - 1 = 0

- 2 sin (x) + 2 sin (2 x) = 0

sin (x) - 8 = cos (x) - 8

What is the general solution for cos(2θ)= 3/5 ?
The general solution for cos(2θ)= 3/5 is θ=(0.92729…)/2+pin,θ=pi-(0.92729…)/2+pin