Finding range of $a$ in exponential inequality

Question

If $a4^{\tan x}+a4^{-\tan x}-2=0$ has a real solution, where $0\leq x\leq \pi,x\neq \frac{\pi}{2},$ then interval of $a$ is

Thoughts on that problem:

Via the arithmetic-geometric inequality (AM-GM), we have

$$4^{\tan x}+4^{-\tan x}\geq 2\implies\frac{2}{a}\geq 2\implies a<1$$

But this is not right

Help me to solve it please

Answer 1

HINT:

We have $4^{\tan x}+4^{-\tan x}=2/a$ so let $u=4^{\tan x}$. Then we get the equation $$u+\frac1u=\frac2a\implies u^2-\frac2au+1=0.$$ After solving for $u$, the value of $x$ can be found by $x=\arctan\log_4 u$ and you can determine the ranges accordingly.

Answer 2

An alternative to TheSimpliFire's approach: $$ \frac{2}{a} = 4^{\tan(x)}+4^{-\tan(x)} = e^{\ln(4)\tan(x)} + e^{-\ln(4)\tan(x)} \Rightarrow \cosh(\ln(4)\tan(x)) = a^{-1}. $$

Now for $x \in [0,\pi]/\tfrac{\pi}{2}$ the range of $\cosh(\ln(4)\tan(x))$ is $[1,\infty)$ ...