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Popular >

solvefor θ,r=sqrt(1+cos(2θ))

Solution

solve for

θ, r = 1 + cos (2 θ)

Solution

θ = \frac{arccos ( r ^{2} - 1 )}{2} + πn, θ = - \frac{arccos ( r ^{2} - 1 )}{2} + πn

Solution steps

r = 1 + cos (2 θ)

Switch sides

1 + cos (2 θ) = r

Square both sides:

1 + cos (2 θ) = r^{2}

1 + cos (2 θ) = r

(1 + cos (2 θ))^{2} = r^{2}

Expand

(1 + cos (2 θ))^{2} : 1 + cos (2 θ)

(1 + cos (2 θ))^{2}

Apply radical rule:

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

= ((1 + cos (2 θ))^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= (1 + 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

= 1 + cos (2 θ)

1 + cos (2 θ) = r^{2}

1 + cos (2 θ) = r^{2}

Solve

1 + cos (2 θ) = r^{2} : cos (2 θ) = r^{2} - 1

1 + cos (2 θ) = r^{2}

Move

1

to the right side

1 + cos (2 θ) = r^{2}

Subtract

1

from both sides

1 + cos (2 θ) - 1 = r^{2} - 1

Simplify

cos (2 θ) = r^{2} - 1

cos (2 θ) = r^{2} - 1

cos (2 θ) = r^{2} - 1

Verify Solutions:

cos (2 θ) = r^{2} - 1 {r \geq 0}

Check the solutions by plugging them into

1 + cos (2 θ) = r

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

Plug

cos (2 θ) = r^{2} - 1 : 1 + r^{2} - 1 = r \Rightarrow r \geq 0

1 + (r^{2} - 1) = r

Square both sides:

r^{2} = r^{2}

1 + r^{2} - 1 = r

(1 + r^{2} - 1)^{2} = r^{2}

Expand

(1 + r^{2} - 1)^{2} : r^{2}

(1 + r^{2} - 1)^{2}

Apply radical rule:

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

= ((1 + r^{2} - 1)^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= (1 + r^{2} - 1)^{\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

= 1 + r^{2} - 1

Refine

= r^{2}

r^{2} = r^{2}

r^{2} = r^{2}

Both sides are equal

True for all r

Verify Solutions:

r < 0

False

, r = 0

True

, r > 0

True

1 + (r^{2} - 1) = r

Combine domain interval with solution interval:

True for all r

Find the function intervals:

r < 0, r = 0, r > 0

1 + (r^{2} - 1) = r

Find the even roots arguments zeroes:

Solve

1 + (r^{2} - 1) = 0 : r = 0

1 + (r^{2} - 1) = 0

Refine

r^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

r = 0

r = 0

The intervals are defined around the zeroes:

r < 0, r = 0, r > 0

Combine intervals with domain

r < 0, r = 0, r > 0

Check the solutions by plugging them into

1 + (r^{2} - 1) = r

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

Plug

r < 0 : 1 + (r^{2} - 1) \neq = r \Rightarrow

False

The solution is

r \geq 0

The solution is

cos (2 θ) = r^{2} - 1 {r \geq 0}

Apply trig inverse properties

cos (2 θ) = r^{2} - 1

General solutions for

cos (2 θ) = r^{2} - 1

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

2 θ = arccos (r^{2} - 1) + 2 πn, 2 θ = - arccos (r^{2} - 1) + 2 πn

2 θ = arccos (r^{2} - 1) + 2 πn, 2 θ = - arccos (r^{2} - 1) + 2 πn

Solve

2 θ = arccos (r^{2} - 1) + 2 πn : θ = \frac{arccos ( r ^{2} - 1 )}{2} + πn

2 θ = arccos (r^{2} - 1) + 2 πn

Divide both sides by

2

2 θ = arccos (r^{2} - 1) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arccos ( r ^{2} - 1 )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{arccos ( r ^{2} - 1 )}{2} + πn

θ = \frac{arccos ( r ^{2} - 1 )}{2} + πn

Solve

2 θ = - arccos (r^{2} - 1) + 2 πn : θ = - \frac{arccos ( r ^{2} - 1 )}{2} + πn

2 θ = - arccos (r^{2} - 1) + 2 πn

Divide both sides by

2

2 θ = - arccos (r^{2} - 1) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = - \frac{arccos ( r ^{2} - 1 )}{2} + \frac{2 πn}{2}

Simplify

θ = - \frac{arccos ( r ^{2} - 1 )}{2} + πn

θ = - \frac{arccos ( r ^{2} - 1 )}{2} + πn

θ = \frac{arccos ( r ^{2} - 1 )}{2} + πn, θ = - \frac{arccos ( r ^{2} - 1 )}{2} + πn

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