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Popular >

solvefor θ,y=-2+cos(3θ)

Solution

solve for

θ, y = - 2 + cos (3 θ)

Solution

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

Solution steps

y = - 2 + cos (3 θ)

Switch sides

- 2 + cos (3 θ) = y

Move

2

to the right side

- 2 + cos (3 θ) = y

Add

2

to both sides

- 2 + cos (3 θ) + 2 = y + 2

Simplify

cos (3 θ) = y + 2

cos (3 θ) = y + 2

Apply trig inverse properties

cos (3 θ) = y + 2

General solutions for

cos (3 θ) = y + 2

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

3 θ = arccos (y + 2) + 2 πn, 3 θ = - arccos (y + 2) + 2 πn

3 θ = arccos (y + 2) + 2 πn, 3 θ = - arccos (y + 2) + 2 πn

Solve

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

3 θ = arccos (y + 2) + 2 πn

Divide both sides by

3

3 θ = arccos (y + 2) + 2 πn

Divide both sides by

3

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

Simplify

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

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

Solve

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

3 θ = - arccos (y + 2) + 2 πn

Divide both sides by

3

3 θ = - arccos (y + 2) + 2 πn

Divide both sides by

3

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

Simplify

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

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

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

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