How to solve the following equation? $$\tan x= \tan(x+10^\circ)\tan(x+20^\circ)\tan(x+30^\circ)$$
Find the acute angle $x$ for $\tan x = \tan(x+10^\circ)\tan(x+20^\circ)\tan(x+30^\circ)$.
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5Do you mean acute? – 2012-12-25
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0Er,yeah, sorry~ – 2012-12-26
1 Answers
$$\frac{\sin x\cos(x+10)}{\cos x\sin(x+10)}=\frac{\sin(x+20)\sin(x+30)}{\cos(x+20)\cos(x+30)}$$
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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0Complete answer. Nice + – 2012-12-25
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1@BabakSorouh, I've now rectified it & kept it a little incomplete as it's after all a homework. – 2012-12-25
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0A little error:The first equation may be $\frac{\sin x\cos(x+10)}{\cos x\sin(x+10)}=\frac{\sin(x+20)\sin(x+30)}{\cos(x+20)\cos(x+30)}$ – 2012-12-26
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0@tan9p, sorry for the typo. Rectified. – 2012-12-26