Maliyah Robles

Answered

2022-07-11

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

Answer & Explanation

Charlee Gentry

Expert

2022-07-12Added 19 answers

hint is: $a\mathrm{cos}\theta +b\mathrm{sin}\theta =c$

Given: $(\sqrt{3}-1)\mathrm{cos}\theta +(\sqrt{3}+1)\mathrm{sin}\theta =2$

Let $(\sqrt{3}-1)=r\mathrm{cos}\alpha $ and $(\sqrt{3}+1)=r\mathrm{sin}\alpha $

Then $r\mathrm{cos}\alpha \mathrm{cos}\theta +r\mathrm{sin}\alpha \mathrm{sin}\theta =2\Rightarrow r\mathrm{cos}(\theta -\alpha )=2\Rightarrow \mathrm{cos}(\theta -\alpha )=\frac{2}{r}$

Now, $r=\sqrt{(\sqrt{3}-1{)}^{2}+(\sqrt{3}+1{)}^{2}}=\sqrt{8}=2\sqrt{2}$

Thus, $\mathrm{cos}(\theta -\alpha )=\frac{1}{\sqrt{2}}=\mathrm{cos}\frac{\pi}{4}$

Also, $\mathrm{tan}\alpha =\frac{\sqrt{3}+1}{\sqrt{3}-1}=\mathrm{tan}(\frac{\pi}{2}-\frac{\pi}{3}+\frac{\pi}{4})\Rightarrow \alpha =\frac{5\pi}{12}$

Thus: $(\theta -\alpha )=2n\pi \pm \frac{\pi}{4}$. Giving, $\theta =2n\pi \pm \frac{\pi}{4}+\frac{5\pi}{12}.$

Given: $(\sqrt{3}-1)\mathrm{cos}\theta +(\sqrt{3}+1)\mathrm{sin}\theta =2$

Let $(\sqrt{3}-1)=r\mathrm{cos}\alpha $ and $(\sqrt{3}+1)=r\mathrm{sin}\alpha $

Then $r\mathrm{cos}\alpha \mathrm{cos}\theta +r\mathrm{sin}\alpha \mathrm{sin}\theta =2\Rightarrow r\mathrm{cos}(\theta -\alpha )=2\Rightarrow \mathrm{cos}(\theta -\alpha )=\frac{2}{r}$

Now, $r=\sqrt{(\sqrt{3}-1{)}^{2}+(\sqrt{3}+1{)}^{2}}=\sqrt{8}=2\sqrt{2}$

Thus, $\mathrm{cos}(\theta -\alpha )=\frac{1}{\sqrt{2}}=\mathrm{cos}\frac{\pi}{4}$

Also, $\mathrm{tan}\alpha =\frac{\sqrt{3}+1}{\sqrt{3}-1}=\mathrm{tan}(\frac{\pi}{2}-\frac{\pi}{3}+\frac{\pi}{4})\Rightarrow \alpha =\frac{5\pi}{12}$

Thus: $(\theta -\alpha )=2n\pi \pm \frac{\pi}{4}$. Giving, $\theta =2n\pi \pm \frac{\pi}{4}+\frac{5\pi}{12}.$

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