How to solve this trigonomectric equation $2^{\lvert\sin x\rvert} + 2^{\lvert\cos x\rvert} = 3$? I thought its solutions are $(\sin x;\cos x) = \{(\pm1;0),(0;\pm 1)\}$
But how can I solve it?
Please help :(
Give the periodicity and symmetries of the trig functions, it suffices to show that $f(u)=2^u+2^{\sqrt{1-u^2}}$ is increasing for $0\lt u\le1/\sqrt2$.  That will imply that the only solutions to $2^{|\sin x|}+2^{|\cos x|}=3$ occur at multiples of $\pi/2$.
We have
$$f'(u)=\ln2\left(2^u-{u\over\sqrt{1-u^2}}2^{\sqrt{1-u^2}}\right)$$
so it's enough to show that
$${2^u\over u}\gt{2^{\sqrt{1-u^2}}\over\sqrt{1-u^2}}$$
for $0\lt u\lt1/\sqrt2$. But notice that in this range we actually have $0\lt u\lt1/\sqrt2\lt\sqrt{1-u^2}\lt1$ (i.e., $0\lt\sin x\lt1/\sqrt2\lt\cos x\lt1$ for $0\lt x\lt\pi/4$). So we need only prove that the function $g(u)=2^u/u$ is decreasing for $0\lt u\lt1$. This is equivalent to showing that $h(u)=(\ln2)u-\ln u$ is decreasing on the same range, which is clear since
$$h'(u)=\ln2-{1\over u}\lt0\iff0\lt u\lt{1\over\ln 2}$$
and $\ln2\lt1$.
For simplicity we have to solve for $x\ge 0$ $$f(x)=2^x+2^{\sqrt{1-x^2}}=3$$.
Note that $$f'(x)=2^x\log 2-\frac{2^{\sqrt{1-x^2}}x\log 2}{\sqrt{1-x^2}}$$
if   $x\le \frac{1}{\sqrt{2}}$ then $x\le \sqrt{1-x^2}$
then we have
Lemma: $$\frac{2^x}{x}\ge \frac{2^{\sqrt{1-x^2}}}{\sqrt{1-x^2}}$$
proof: show that $g(x)=\frac{2^x}{x}$ is decreasing in given interval.
From this lemma it follows that $f'(x)\ge 0$.Thus for all $x\in (0,\frac{1}{\sqrt{2}}]$ :$f(x)&gt;f(0)=3$
Similarly prove that $f(x)$ is decreasing for all $x\in [\frac{1}{\sqrt{2}},1]$ hence deduce that $f(x)&gt;f(1)=3$ for all $x\in [\frac{1}{\sqrt{2}},1)$.
The conclusion is now obvious:i.e $x=0,x=1$ are the only solutions.