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

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

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1=sqrt(4cos(2θ))

1 = 4 cos (2 θ)

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

Solution steps

1 = 4 cos (2 θ)

Switch sides

4 cos (2 θ) = 1

Square both sides:

4 cos (2 θ) = 1

4 cos (2 θ) = 1

Solve

4 cos (2 θ) = 1 : cos (2 θ) = \frac{1}{4}

cos (2 θ) = \frac{1}{4}

Verify Solutions:

cos (2 θ) = \frac{1}{4}

True

The solution is

cos (2 θ) = \frac{1}{4}

Apply trig inverse properties

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

Solve

2 θ = 2 π - 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

209 cos (4 7^{\circ}) = 150 cos (θ)

7 - 5 cos (x) = 10, 7 - 5 cos (x) = 10

cos (4 x) = cos (x)

- 4 cos (2 θ) - 19 = - 26 cos (θ), 0 \leq θ < 36 0^{\circ}

3 cos^{2} (x) = 7 - 8 sin (x)

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