What is the general solution for cos(2θ)= 1/4 ?

The general solution for cos(2θ)= 1/4 is θ=(1.31811…)/2+pin,θ=pi-(1.31811…)/2+pin

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cos(2θ)= 1/4

cos (2 θ) = \frac{1}{4}

θ = \frac{1.31811 \dots}{2} + πn, θ = π - \frac{1.31811 \dots}{2} + πn

Degrees

θ = 37.76124 \dots^{\circ} + 18 0^{\circ} n, θ = 142.23875 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 θ) = \frac{1}{4}

Apply trig inverse properties

cos (2 θ) = \frac{1}{4}

General solutions for

cos (2 θ) = \frac{1}{4}

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

2 θ = arccos (\frac{1}{4}) + 2 πn, 2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

2 θ = arccos (\frac{1}{4}) + 2 πn, 2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

Solve

2 θ = arccos (\frac{1}{4}) + 2 πn : θ = \frac{arccos ( \frac{1}{4} )}{2} + πn

2 θ = arccos (\frac{1}{4}) + 2 πn

Divide both sides by

2

2 θ = arccos (\frac{1}{4}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{arccos ( \frac{1}{4} )}{2} + πn

θ = \frac{arccos ( \frac{1}{4} )}{2} + πn

Solve

2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn : θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

Divide both sides by

2

2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

Simplify

θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

θ = \frac{arccos ( \frac{1}{4} )}{2} + πn, θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

Show solutions in decimal form

θ = \frac{1.31811 \dots}{2} + πn, θ = π - \frac{1.31811 \dots}{2} + πn

sec (θ) = \frac{25}{7}

(3 + sin (x))^{2} = 13 - cos^{2} (x)

sec (x) = 3, \frac{3 π}{2} \leq x \leq 2 π

tan (2 x) = cot (3 x)

sin (x) = \infty

What is the general solution for cos(2θ)= 1/4 ?
The general solution for cos(2θ)= 1/4 is θ=(1.31811…)/2+pin,θ=pi-(1.31811…)/2+pin