# Find the acute angle $x$ for $\tan x = \tan(x+10^\circ)\tan(x+20^\circ)\tan(x+30^\circ)$.

How to solve the following equation? $$\tan x= \tan(x+10^\circ)\tan(x+20^\circ)\tan(x+30^\circ)$$

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Do you mean acute? –  Nameless Dec 25 '12 at 12:29
Er,yeah, sorry~ –  tan9p Dec 26 '12 at 11:36

$$\frac{\sin x\cos(x+10)}{\cos x\sin(x+10)}=\frac{\sin(x+20)\sin(x+30)}{\cos(x+20)\cos(x+30)}$$

Applying componendo and dividendo,

$$\frac{\cos x\sin(x+10)+\sin x\cos(x+10)}{\cos x\sin(x+10)-\sin x\cos(x+10)} =\frac{\cos(x+20)\cos(x+30)+\sin(x+20)\cos(x+30)}{\cos(x+20)\cos(x+30)-\sin(x+20)\cos(x+30)}$$

$$\frac{\sin(2x+10)}{\sin10}=\frac{\cos 10}{\cos(2x+50)}$$ applying $\sin(A\pm B)$ and $\cos(A\pm B)$

So, $$\sin(2x+10)\cos(2x+50)=\sin10 \cos 10$$

$$\sin(4x+60)-\sin40=\sin20$$ (applying $2\sin A\cos A=\sin2A$ and $2\sin A\cos B==\sin(A+B)+\sin(A-B)$)

$$\sin(4x+60)=\sin40+\sin20=2\sin\frac{20+40}2\cos\frac{40-20}2=\cos10$$

(applying $\sin 2C+\sin 2D =2\sin(C+D)\cos(C-D)$ and $\sin(90\pm A)=\cos A$)

Now, $\sin(4x+60)=\cos\{90-(4x+60)\}=\cos(30-4x)=\cos(4x-30)$ as $\cos(-A)=\cos A$

So, $\cos(4x-30)=\cos10$

or, $4x-30=360n\pm10$ where $n$ is any integer.

Find suitable $n$ to keep $x\in[0,90]$

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Complete answer. Nice + –  Babak S. Dec 25 '12 at 13:35
@BabakSorouh, I've now rectified it & kept it a little incomplete as it's after all a homework. –  lab bhattacharjee Dec 25 '12 at 16:52
Nice.Thank you. –  tan9p Dec 26 '12 at 11:34
A little error:The first equation may be $\frac{\sin x\cos(x+10)}{\cos x\sin(x+10)}=\frac{\sin(x+20)\sin(x+30)}{\cos(x+20)\cos(x+30)}$ –  tan9p Dec 26 '12 at 11:49
@tan9p, sorry for the typo. Rectified. –  lab bhattacharjee Dec 26 '12 at 11:57