Prove that $\tan x =\sin x + 1$ have only one solution in $(-\frac{\pi}{2},\frac{\pi}{2})$ Prove that $\tan x = \sin x + 1$ have only one solution in $(-\frac{\pi}{2},\frac{\pi}{2})$
Well, I believe I need to show that there's a solution. so I need to find x which solves $\tan x = \sin x + 1$.
Then, I need to derivate it and show that the derivation either bigger than zero, or smaller than zero, Thus - there will be no solutions.
So in intuition way, I know how to solve it, but practically I'm having difficulties. Can you guys help with the steps I've shown?
I believe It's Rolle's rule. Because we just studied it :)
 A: let me rewrite the equation as $$\tan t = \sin t + 1.\tag 1$$  i will use $x = \cos t, y = \sin t.$ then $(1)$ becomes $$\frac y x = y + 1 \to y = \frac x{1-x}\tag 2 $$
the solution to $(1)$ are the points where the hyperbola (2) cuts the unit circle in the first quadrant. this is easy to see because the  branch of the hyperbola going through $(0,0)$ can only cut the unit circle at one in the first quadrant and the second time at a pint in the third quadrant.
A: You can use the formulas
$$
\tan x=\frac{2t}{1-t^2},\qquad
\sin x=\frac{2t}{1+t^2}
$$
where $t=\tan(x/2)$. Then the equation becomes
$$
\frac{2t}{1-t^2}=\frac{2t}{1+t^2}+1
$$
that can be rewritten
$$
2t+2t^3=2t-2t^3+1-t^4
$$
and finally
$$
t^4+4t^3-1=0
$$
We want only solutions in the range $-1<t<1$, because we want $-\pi/2<x<\pi/2$, so $-\pi/4<x/2<\pi/4$.
Set $f(t)=t^4+4t^3-1$; since $f(-1)=-4$ and $f(1)=4$, we know that one solution exists.
Consider $f'(t)=4t^3+12t^2=4t^2(t+3)$. This is positive for $t\in[-1,1]$, except at $0$; therefore the function is strictly increasing in the given interval, hence we have only one solution.

You can also consider
$$
f(x)=\tan x-\sin x-1
$$
and note that
$$
\lim_{x\to-\pi/2^+}f(x)=-\infty,
\qquad
\lim_{x\to\pi/2^-}f(x)=\infty
$$
which implies a solution exists. Then
$$
f'(x)=\frac{1}{\cos^2x}-\cos x=\frac{1-\cos^3x}{\cos^2x}
$$
which only vanishes at $0$ and is positive elsewhere, in the given domain $(-\pi/2,\pi/2)$. Thus $f$ is strictly increasing.
A: $\tan(x)$ is a bijection from $]-\pi/2, \pi/2[$ to $]-\infty, \infty[$ as a continous and increasing function (derivative >0). $\sin(x)+1$ is increasing, continuous function: this is a bijection from $]-\pi/2, \pi/2[$ to $]0,2[$.
hence, the two graphs must have a unique common point as $\sin(x)+1 \in [0, 2|$. To have an interval, you apply the Bolzano's theorem:
if f is continuous on [a,b] and f(a)< 0 and f(b)>0 then, there is $\alpha$ where $f(\alpha)=0$. $\alpha$ is unique if f is a bijection on [a,b] (example : f monotone).
here $f(x)=\tan(x)-\sin(x)-1$. we have $\alpha=1.08264952...$   
