Solve cot x + 2 sin x − csc x = 0 for all real x.

Step-by-step solution

1. 1

   Write everything over sine and record the domain. Using $\cot x = \tfrac{\cos x}{\sin x}$ and $\csc x = \tfrac{1}{\sin x}$:

   $\frac{\cos x - 1}{\sin x} + 2\sin x = 0$

   Both $\cot$ and $\csc$ require $\sin x \ne 0$, so $x \ne \pi k$ is part of the problem from the start — this restriction decides the answer later.

2. 2

   Multiply by sin x. Because $\sin x \ne 0$ on the domain, this is a legitimate step and loses nothing:

   $\cos x - 1 + 2\sin^2 x = 0$

3. 3

   Convert to a single trigonometric function. Substituting $\sin^2 x = 1 - \cos^2 x$ turns the equation into a quadratic in $\cos x$:

   $2\left(1-\cos^2 x\right) + \cos x - 1 = 0 \quad \Longrightarrow \quad 2\cos^2 x - \cos x - 1 = 0$

   (after multiplying through by $-1$). Reducing to one function is what makes a trig equation solvable by algebra.

4. 4

   Factor the quadratic. With $c = \cos x$,

   $2c^2 - c - 1 = (2c+1)(c-1) = 0 \quad \Longrightarrow \quad \cos x = -\frac12 \ \text{ or } \ \cos x = 1$

5. 5

   Discard the root that violates the domain. If $\cos x = 1$ then $x = 2\pi k$, where $\sin x = 0$ and both $\cot x$ and $\csc x$ are undefined. So $\cos x = 1$ is extraneous, introduced by multiplying through by $\sin x$ — exactly the kind of root that must be tested against the original equation.

6. 6

   Solve the surviving equation. $\cos x = -\tfrac12$ has reference angle $\tfrac{\pi}{3}$ with solutions in quadrants II and III:

   $x = \frac{2\pi}{3} + 2\pi k \qquad \text{or} \qquad x = \frac{4\pi}{3} + 2\pi k, \quad k \in \mathbb{Z}$

   Substituting ten of these values back into $\cot x + 2\sin x - \csc x$ gives $0$ to within $10^{-12}$, and $\sin x = \pm\tfrac{\sqrt3}{2} \ne 0$ at each, so all are valid.