Miscellaneous · Q11: Solve $2\tan^{-1}(\cos x)=\tan^{-1}(2\operatorname{cosec}x)$

Basics you need

Solution

1. Where the equation makes sense

   The right side contains $\operatorname{cosec}x$⁠, so we need $\sin x \ne 0$⁠, that is

   $$x \ne n\pi, \qquad n\in\Z.$$

   For such $x$⁠, $\cos^2 x = 1 - \sin^2 x < 1$⁠, so

   $$|\cos x| < 1.$$

   Both $\tan^{-1}$ terms are then defined, since $\tan^{-1}$ accepts every real number.

   Any solution must come from this set, so we find it first. The fact $|\cos x| < 1$ will be exactly what makes the next step legal. Notice that the excluded points $x = n\pi$ are precisely where $\cos x = \pm 1$⁠.
   $\operatorname{cosec} x$ is undefined at $x = n\pi$ (Basic 1); $\sin^2x + \cos^2x = 1$⁠.

2. The double-angle rule for tan⁻¹, and why it needs |t| < 1

   Let $|t| < 1$ and put $\varphi = \tan^{-1}t$⁠. Since $-1 < t < 1$⁠, $\varphi\in\left(-\frac\pi4,\frac\pi4\right)$⁠, so

   $$2\varphi \in \left(-\tfrac\pi2, \tfrac\pi2\right), \qquad \tan 2\varphi = \frac{2t}{1-t^2}.$$

   Because $2\varphi$ lies in the principal range of $\tan^{-1}$⁠, $\tan^{-1}(\tan 2\varphi) = 2\varphi$⁠, that is

   $$2\tan^{-1}t = \tan^{-1}\frac{2t}{1-t^2}, \qquad |t| < 1.$$

   Apply it with $t = \cos x$⁠, allowed by Step 1, and use $1 - \cos^2x = \sin^2x$⁠:

   $$2\tan^{-1}(\cos x) = \tan^{-1}\frac{2\cos x}{\sin^2 x}.$$

   In particular, on the whole domain the left side $2\tan^{-1}(\cos x)$ lies strictly inside $\left(-\frac\pi2,\frac\pi2\right)$⁠: it never reaches $\pm\frac\pi2$⁠, because that would need $\cos x = \pm1$⁠, which happens only at the excluded points $x = n\pi$⁠. So both sides always take values in the range of $\tan^{-1}$⁠, and no solution can be hidden by a range mismatch.

   For $|t| > 1$ the rule is false: $2\varphi$ would leave $\left(-\frac\pi2,\frac\pi2\right)$ and the two sides would differ by $\pi$ (for example $2\tan^{-1}2 \approx 2.214$⁠, but $\tan^{-1}\frac{4}{-3} \approx -0.927$⁠). Using it outside $|t|<1$ can lose or invent solutions. Here every allowed $x$ has $|\cos x| < 1$⁠, so the rewrite is exact on the whole domain: no solution is lost or added by this step.
   $\tan 2\varphi = \frac{2\tan\varphi}{1-\tan^2\varphi}$ (Basic 3) and $\tan^{-1}(\tan\alpha) = \alpha$ for $\alpha\in\left(-\frac\pi2,\frac\pi2\right)$⁠. Lesson: the folding rules

3. Equal tan⁻¹ values mean equal arguments

   The equation is now

   $$\tan^{-1}\frac{2\cos x}{\sin^2x} = \tan^{-1}\frac{2}{\sin x}.$$

   Since $\tan^{-1}$ is one-one, this holds exactly when

   $$\frac{2\cos x}{\sin^2x} = \frac{2}{\sin x}.$$
   A one-one function gives equal outputs only for equal inputs, and equal inputs give equal outputs because $\tan^{-1}$ is a function (one output per input). So this step is an "if and only if": it neither loses nor adds solutions.
   $\tan^{-1}$ is one-one on $\R$ (Basic 2).

4. Clear the fractions and solve

   Multiply both sides by $\frac{\sin^2 x}{2}$⁠, which is not zero because $\sin x \ne 0$⁠:

   $$\cos x = \sin x.$$

   Here $\cos x \ne 0$⁠: if $\cos x$ were 0, then $\sin x = \cos x = 0$⁠, impossible since $\sin^2x + \cos^2 x = 1$⁠. So we may divide by $\cos x$⁠:

   $$\tan x = 1 \iff x = n\pi + \frac\pi4, \quad n\in\Z.$$

   Each of these has $\sin x = \pm\frac{1}{\sqrt2} \ne 0$⁠, so all of them lie in the domain from Step 1.

   Multiplying by a non-zero number and dividing by a non-zero number are both reversible, so again nothing is lost or added. Finally we confirm that the solutions avoid the excluded points $x = n\pi$⁠, where $\operatorname{cosec}x$ is undefined.
   General solution of $\tan x = 1$ (Basic 4).

5. Check every solution in the original equation

   Every step was reversible, so all of $x = n\pi + \frac\pi4$ are solutions and there are no others. Still, check both kinds directly.

   $n$ even, e.g. $x = \frac\pi4$⁠: $\cos x = \frac1{\sqrt2}$⁠, $\operatorname{cosec}x = \sqrt2$⁠.

   $$\begin{aligned} \text{LHS} &= 2\tan^{-1}(0.7071067812) \\ &= 2(0.6154797087) = 1.2309594173, \\ \text{RHS} &= \tan^{-1}(2.8284271247) = 1.2309594173. \end{aligned}$$

   $n$ odd, e.g. $x = \frac{5\pi}4$⁠: $\cos x = -\frac1{\sqrt2}$⁠, $\operatorname{cosec}x = -\sqrt2$⁠, and both sides equal $-1.2309594173$⁠, because $\tan^{-1}$ is odd.

   Adding $2\pi$ changes neither $\cos x$ nor $\operatorname{cosec}x$⁠, so these two cases cover every $n$⁠.

   The figure is a useful sanity check on the claim that there are no other solutions: one crossing in each interval of length $\pi$⁠, at $\frac\pi4$ past each excluded point. At a non-solution such as $x = \frac\pi2$ the sides differ: the left is $2\tan^{-1}0 = 0$⁠, the right is $\tan^{-1}2 = 1.1071487178$⁠.
   $\tan^{-1}(-a) = -\tan^{-1}a$⁠; $\cos$ and $\operatorname{cosec}$ have period $2\pi$⁠.

Every fact used, and where it comes from