How can I represent the root of $\sin x=\cos x+\tan x$ as “length”?

Question

Quesiton:
Represent the root(s) of $\sin x=\cos x+\tan x$ as length on rectangular coordinate.

For example, if $x=2$, you represent it as "the length between $(0,0)$ and $(2,0)$".

How can I solve this?

Answer 1

The equation $\sin x=\cos x+\tan x$ is equivalent to

$$\frac{2\tan \frac{x}{2}}{1+\tan ^{2}\frac{x}{2}}=\frac{1-\tan ^{2}\frac{x}{2% }}{1+\tan ^{2}\frac{x}{2}}+\frac{2\tan \frac{x}{2}}{1-\tan ^{2}\frac{x}{2}}.$$

Set $\tan \frac{x}{2}=y$. Then

$$2y=1-y^{2}+\left( 2y\right) \frac{1+y^{2}}{1-y^{2}},$$

or, equivalently

$$y^{4}+4y^{3}-2y^{2}+1=0.$$

Then $x=2\arctan y$, where $y$ are the solutions of this quartic (see this computation in Wolfram Alpha).

The direct computation in Wolfram Alpha gives solution(s) in terms of $\arccos(R(x))$ where $R(x)$ is a function with too many radicals.

Answer 2

I think I found another method to solve this problem. Let me know any errors of this.

There's an unit circle which center is (0, 0). And A(x', y') is on this circle.
Then we know $\sin x=y'$, $\cos x=x'$, and $\tan x= \frac{y'}{x'}$.

$$y' = x' + \frac{x'}{y'}$$
$$x'y' = (x')^2 + y'$$
$$(x'-1)y' = (x')^2$$
So, $y' = x'+1+\frac{1}{x'-1}$. And $(x')^2+(y')^2=1$.

We can get the intersection points of these.
Start at (1, 0), follow the circle's circumference, and end at that intersection.
The length of that curve is same with the roots of $\sin x = \cos x + \tan x$.

(I checked with calculator and these links: (1) , (2) )