Finding the number of solutions for $e^{\tan x}=\sin x+ \cos x$ graphically

Question

The number of solutions of the equation $$e^{\tan x}=\sin x+ \cos x$$ In the interval $[-\pi,\pi]$

I tried to approach this question graphically but hit a dead end when trying to graph $e^{tanx}$

For graphing sinx+cosx I graphed $\sqrt 2(sin(x+\frac{\pi}{4}))$

I still feel this question can be solved graphically,Is there a method to graph $e^{tanx}$ to some accuracy ?

Answer 1

I'd make a few observations about $f(x) =e^{\tan x}$.

First, it is always positive.

Second, it is periodic, so if you can sketch it for $0 < x < \pi$ then you can repeat that for $-\pi < x < 0$

Third, let's think about what happens when $x$ is $-\pi$, $- \pi/2$, $0$, $\pi / 2$, $\pi$:

$f(-\pi) = f(0) = f(\pi) = e^0 = 1$

For $x <\pi/2$ as $x\to \pi/2, f(x)\to \infty$, while for $x>\pi/2$ as $x\to \pi/2, f(x)\to 0$. Same behaviour around $-\pi/2$.

Fourth, $f(\pi/4) = e$, $f(3\pi/4) = e^{-1}$.

So, to sketch, start at $(0,1)$ and draw a curve through $(\pi/4,e)$ that shoots off to infinity at $\pi/2$.

Then sketch a curve from $(\pi/2,0)$ through $(3\pi/4,e^{-1})$ to $(\pi,1)$.

Answer 2

There is the trivial solution $x=0$.

For tho other root start plotting from $$x=\frac \pi 2 +\text{some small number}$$ to $x=\pi$. Now, zoom more and more.

Answer 3

As you can see from the graph below there are two roots in the given interval $[-\pi,\pi]$.

Given $$f(x)=e^{\tan x}-\sin x-\cos x$$ One root is quite obvious since $f(0)=0$.

To find the other we use the Newton method. Compute the derivative $$f'(x)=\sin x-\cos x+e^{\tan x} \sec ^2 x$$

starting from $x_0=2$ we define $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

$$ \begin{array}{c|r|r} n & x_n & \text{error}\\ \hline 0 & 2 & - \\ 1 & 2.1487490 & 0.851251 \\ 2 & 2.1845423 & 0.0357933 \\ 3 & 2.1844057 & 0.000136542 \\ 4 & 2.1844057 & 1.625\times 10^{-9} \\ \end{array} $$

$$ ... $$

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