What is the general solution for cot(θ)=(1+cos^2(θ))/(2sin(θ)cos(θ)) ?

The general solution for cot(θ)=(1+cos^2(θ))/(2sin(θ)cos(θ)) is No Solution for θ\in\mathbb{R}

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cot(θ)=(1+cos^2(θ))/(2sin(θ)cos(θ))

cot (θ) = \frac{1 + cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )}

No Solution for θ \in R

Solution steps

cot (θ) = \frac{1 + cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )}

Subtract

\frac{1 + cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )}

from both sides

cot (θ) - \frac{1 + cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )} = 0

Simplify

cot (θ) - \frac{1 + cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )} : \frac{2 cot ( θ ) sin ( θ ) cos ( θ ) - 1 - cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )}

\frac{2 cot ( θ ) sin ( θ ) cos ( θ ) - 1 - cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 cot (θ) sin (θ) cos (θ) - 1 - cos^{2} (θ) = 0

Rewrite using trig identities

- 1 + cos^{2} (θ) = 0

Solve by substitution

cos (θ) = 1, cos (θ) = - 1

cos (θ) = 1 : θ = 2 πn

cos (θ) = - 1 : θ = π + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

Since the equation is undefined for:

2 πn, π + 2 πn

No Solution for θ \in R

cot (x) = 3 + 1

sec (3 x) - csc (3 0^{\circ}) = 0, \frac{x + 35}{5}

4 cos (2 θ) - 10 cos (θ) + 14 = 7, 0 \leq θ < 36 0^{\circ}

x, Z = sin (x y^{2})

2 tan^{2} (θ) = 2

What is the general solution for cot(θ)=(1+cos^2(θ))/(2sin(θ)cos(θ)) ?
The general solution for cot(θ)=(1+cos^2(θ))/(2sin(θ)cos(θ)) is No Solution for θ\in\mathbb{R}