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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?

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    Why would "the length between $(0,0)$ and $(u,0)$, where u is *a* solution to $\sin\,u=\cos\,u+\tan\,u$ not suffice?"2011-07-23
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    That's right. But I want to know the exact value of root, and I want to represent it as "length".2011-07-23
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    Generally there is no reason to expect a simple expression for a transcendental equation such as yours... but FWIW: express everything in terms of either sine or cosine, solve the resulting algebraic equation for sine or cosine, and then take the arcsine/arccosine of that...2011-07-23
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    Thanks. I asked this just for fun. I saw this problem on internet.2011-07-23

1 Answers 1

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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.

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    Of course, pick the appropriate solutions of the quartic, and then remember that you get an infinite number of solutions due to periodicity...2011-07-23