Trigonometric Equation - can this be solved using complex numbers

Question

How do I solve the following:

$\cos (12x) = 5 \sin (3x) + 9 \tan^2( x )+ \cot ^2 (x)$ for $x \in (0,360)$

I tried converting cos and sin term into single angle i.e. into x but the equation becomes messy. Any elegant solution at hand?

Answer 1

The minimum value of $9 \tan^2 x + \cot^2 x$ is $2 \sqrt{9 \tan^2 x \cot^2 x} = 6$ by AM-GM for all real $x$, as $u^2 ≥ 0, u \in \mathbb R$. Thus $\cos(12x) - \left(5 \sin(3x) + 9 \tan^2 x + \cot^2 x \right)$ is bounded above by $g(x)=\cos(12x) - \left(5 \sin(3x) + 6 \right)$.

Now $g(x)$ is again bounded above by $1 -5 (-1) - 6 = 0$. Hence you just need to solve $g(x) = 0$, but as $\cos(12x) ≤ 1$, $5 \sin(3x) + 6 ≥ 1$ for all $x \in \mathbb R$, this implies:

$$\cos(12x) = 1, 5 \sin(3x) + 6 = 1$$

Answer 2

$$a+b\geq 2\sqrt{ab}\\9tan^2x+1cot^2x\geq2\sqrt{9tan^2x\times 1cot^2x}=6$$so $$cos(12x)-5sin(3x)\geq 6 $$Implicit $$cos(12x)-5sin(3x)\leq \max\{cos(12x)\}+\max \{ -5sin(3x)\}\leq 1+5$$ and only possibilities : $$\begin{cases}cos(12x)=1\\sin(3x)=-1\end{cases}$$

Answer 3

Not much different form Toby Mak's answer but We could infact rewrite the given equaltion in the following form $$1-\cos 12x+5(\sin 3x+1)+{\left( 3\tan x-\cot x\right)}^2=0$$ as each terms is $\ge 0$ we must have that each term must be zero ,

can you take it from here?