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I am trying to form a trig general solution for this system of equations: $${ \sin(x)=\frac{\sqrt{3}}{2} , \cos(x)=-\frac{1}{2}}.$$

I found out that the general solution is $$x=2\pi n -\frac{4\pi}{3}\ $$ I did this by solving each equation seperately, finding the values of x that satisfy both, and then working out the general solution by looking at the pattern. However I want to find a way of doing it algebraically.

I tried using the identity $$\frac{\sin(x)}{\cos(x)}=\tan(x)$$ Giving me one equation: $$\tan(x)=-\sqrt{3}.$$

Then I found the general solution for this equation $$x=\pi n-\frac{\pi }{3}$$ However this equation is not the correct general solution - whilst it does encompass all of the correct solutions, it also gives incorrect solutions (the incorrect solutions lie when even values of n are subbed in).

Can anybody help me find a foolproof method for finding a general solution for a pair of trig equations?

Thanks.

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  • $\begingroup$ "It also gives incorrect solutions (the incorrect solutions lie when even values of n are subbed in)." Then force $n$ to be odd, by substituting $n$ with $2n+1$. $\endgroup$
    – Steven
    Sep 21, 2015 at 20:35
  • $\begingroup$ @StevenFontaine Thanks, but will that work for all pairs of trig equations? Do you know why it only holds true for odd values of n? $\endgroup$ Sep 21, 2015 at 21:05

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For this special case you can write:

$$z=\cos(x)+i\sin(x)=-\frac{1}{2}+i\frac{\sqrt{3}}{2}=e^{i(\frac{2}{3}\pi+2\pi k)}$$

Note that $\frac{2}{3}\pi+2\pi k=-\frac{4}{3}\pi+2\pi k'$.

You instantly get all solutions.

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