solvefor θ,r=sqrt(1+cos(2θ))

Solution

solve for

θ, r = 1 + cos (2 θ)

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^{2}

Solve

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

cos (2 θ) = r^{2} - 1

Verify Solutions:

cos (2 θ) = r^{2} - 1 {r \geq 0}

The solution is

cos (2 θ) = r^{2} - 1 {r \geq 0}

Apply trig inverse properties

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

Solve

2 θ = - 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

tan (x) + tan (x + 4 5^{\circ}) = - 2

tan (x) = \frac{22.5}{34}

y, arccot (y) = ln (sin^{2} (3 x)) - e^{x}

tan^{2} (4 x) = 0

cos (\frac{x}{2}) = - 0.4