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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Popular >

1=sqrt(4cos(2θ))

Solution

1 = 4 cos (2 θ)

Solution

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

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

(4 cos (2 θ))^{2} = 1^{2}

Expand

(4 cos (2 θ))^{2} : 4 cos (2 θ)

(4 cos (2 θ))^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((4 cos (2 θ))^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (4 cos (2 θ))^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 4 cos (2 θ)

Expand

1^{2} : 1

1^{2}

Apply rule

1^{a} = 1

= 1

4 cos (2 θ) = 1

4 cos (2 θ) = 1

Solve

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

4 cos (2 θ) = 1

Divide both sides by

4

4 cos (2 θ) = 1

Divide both sides by

4

\frac{4 cos ( 2 θ )}{4} = \frac{1}{4}

Simplify

cos (2 θ) = \frac{1}{4}

cos (2 θ) = \frac{1}{4}

cos (2 θ) = \frac{1}{4}

Verify Solutions:

cos (2 θ) = \frac{1}{4}

True

Check the solutions by plugging them into

4 cos (2 θ) = 1

Remove the ones that don't agree with the equation.

Plug in

cos (2 θ) = \frac{1}{4} :

True

4 (\frac{1}{4}) = 1

4 (\frac{1}{4}) = 1

4 (\frac{1}{4})

Remove parentheses:

(a) = a

= 4 \cdot \frac{1}{4}

Multiply

4 \cdot \frac{1}{4} : 1

4 \cdot \frac{1}{4}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4}{4}

Cancel the common factor:

4

= 1

= 1

Apply rule

1 = 1

= 1

1 = 1

True

The solution is

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

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Frequently Asked Questions (FAQ)

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