Maliyah Robles

2022-07-11

General solution to $\left(\sqrt{3}-1\right)\mathrm{cos}x+\left(\sqrt{3}+1\right)\mathrm{sin}x=2$

Charlee Gentry

Expert

hint is: $a\mathrm{cos}\theta +b\mathrm{sin}\theta =c$
Given: $\left(\sqrt{3}-1\right)\mathrm{cos}\theta +\left(\sqrt{3}+1\right)\mathrm{sin}\theta =2$
Let $\left(\sqrt{3}-1\right)=r\mathrm{cos}\alpha$ and $\left(\sqrt{3}+1\right)=r\mathrm{sin}\alpha$
Then $r\mathrm{cos}\alpha \mathrm{cos}\theta +r\mathrm{sin}\alpha \mathrm{sin}\theta =2⇒r\mathrm{cos}\left(\theta -\alpha \right)=2⇒\mathrm{cos}\left(\theta -\alpha \right)=\frac{2}{r}$
Now, $r=\sqrt{\left(\sqrt{3}-1{\right)}^{2}+\left(\sqrt{3}+1{\right)}^{2}}=\sqrt{8}=2\sqrt{2}$
Thus, $\mathrm{cos}\left(\theta -\alpha \right)=\frac{1}{\sqrt{2}}=\mathrm{cos}\frac{\pi }{4}$
Also, $\mathrm{tan}\alpha =\frac{\sqrt{3}+1}{\sqrt{3}-1}=\mathrm{tan}\left(\frac{\pi }{2}-\frac{\pi }{3}+\frac{\pi }{4}\right)⇒\alpha =\frac{5\pi }{12}$
Thus: $\left(\theta -\alpha \right)=2n\pi ±\frac{\pi }{4}$. Giving, $\theta =2n\pi ±\frac{\pi }{4}+\frac{5\pi }{12}.$

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