solvefor θ,y=-2+cos(3θ)

Solution

solve for

θ, y = - 2 + cos (3 θ)

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

cos (3 θ) = y + 2

Apply trig inverse properties

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}

Solve

3 θ = - arccos (y + 2) + 2 πn : 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}

cos (2 x) + 5 \cdot cos (x) = - 3

2 cos^{2} (x) - sin (x) - 2 = 0

8 sin (x) + 7 cos (x) = 1

sec (y) = 12

- sin (x) + 3 cos (x - \frac{π}{2}) = 1