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  • Find the general solution of the equation (√3 – 1) cos θ + (√3 + 1) sin θ = 2 [Hint: Put √3 – 1 = r sin α, √3 + 1 = r cos α which gives tan α = tan((π/4) – (π/6)) α = π/12]

Find the general solution of the equation (\sqrt 3 - 1) \cos \theta + ( \sqrt 3 + 1) \sin \theta = 2

[Hint: Put Put \: \: \sqrt 3 - 1 = r \sin \alpha , \sqrt 3 + 1 = r \cos \alpha which gives \tan \alpha = \tan(( \pi /4) - ( \pi /6)) \alpha = \pi /12].

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\\\left( \sqrt {3} - 1 \right) \cos \theta + \left( \sqrt {3}+1 \right) \sin \theta =2 \\\\ \text{~ put }\sqrt {3} - 1=r\sin \alpha ,~ \sqrt {3}+1=r\cos \alpha \\\\ \text{~squaring~and adding we get }r^{2}=3+1 - 2\sqrt {3}+3+1+2\sqrt {3}~~ \\\\ ~~r^{2}=8,~~ r=2\sqrt {2}~~ \\\\


  \\ ~~ r\sin \alpha \cos \theta +r\cos \alpha \sin \theta =2 \\\\ \text{~ r} \left( \sin \alpha \cos \theta +\cos \alpha \sin \theta \right) =2 \\\\ ~2\sqrt {2}\sin \left( \alpha + \theta \right) =2~ \\\\ \sin \left( \alpha + \theta \right) =\frac{2}{2\sqrt {2}} \\\\ ~\sin \left( \theta + \alpha \right) =\sin \frac{ \pi }{4}~~ \\\\


  \\ \alpha + \theta =n \pi + \left( - 1 \right) ^{n}\frac{ \pi }{4} \ldots \ldots . \left( i \right) ~ \\\\ \text{~ Now, }\frac{r\sin \theta }{r\cos \alpha }=\frac{\sqrt {3} - 1}{\sqrt {3}+1}~~ \\\\ \tan \alpha =\frac{\tan \frac{ \pi }{3} - \tan \frac{ \pi }{4}}{1+\tan \frac{ \pi }{4}\tan \frac{ \pi }{3}}~ \\\\ ~\tan \alpha =\tan \left( \frac{ \pi }{3} - \frac{ \pi }{4} \right) \\\\ ~~\tan \alpha =\tan \frac{ \pi }{12} \\\\


    \\ \alpha =\frac{ \pi }{12}~ \\\\ \text{ Putting the value of } \alpha { in equation} \left( i \right) \text{~we get }\frac{ \pi }{12}+ \theta =n \pi + \left( - 1 \right) ^{n}.\frac{ \pi }{4}~~~~ \\\\ ~~~~~ \theta =n \pi + \left( - 1 \right) ^{n}.\frac{ \pi }{4} - \frac{ \pi }{12}~ \\\\

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