JEE Main & Advanced Mathematics Trigonometric Equations General Solution of The Form \[\mathbf{acos}\,\mathbf{\theta +bsin}\,\mathbf{\theta =c}\]

General Solution of The Form \[\mathbf{acos}\,\mathbf{\theta +bsin}\,\mathbf{\theta =c}\]

Category : JEE Main & Advanced

In \[a\cos \theta \,+b\sin \theta =c,\] put \[a=r\,\cos \alpha \] and \[b=r\,\sin \alpha \]where \[r=\sqrt{{{a}^{2}}+{{b}^{2}}}\] and \[|c|\le \sqrt{{{a}^{2}}+{{b}^{2}}}\]

 

Then,\[r\,(\cos \alpha \,\cos \theta +\sin \alpha \,\sin \theta )=c\]

 

\[\Rightarrow \,\,\cos (\theta -\alpha )=\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=\cos \beta \], (say)                      .....(i)

 

\[\Rightarrow \,\,\theta -\alpha =2n\pi \pm \beta \Rightarrow \,\theta =2n\pi \pm \beta +\alpha ,\] where \[\tan \alpha =\frac{b}{a}\] is the general solution.

 

Alternatively, putting \[a=r\,\sin \alpha \] and \[b=r\,\cos \alpha \],

 

where \[r=\sqrt{{{a}^{2}}+{{b}^{2}}}\] \[\Rightarrow \,\,\sin (\theta +\alpha )=\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=\sin \gamma \], (say)

 

\[\Rightarrow \,\,\theta +\alpha =n\pi +{{(-1)}^{n}}\gamma \]\[\Rightarrow \,\,\theta =n\pi +{{(-1)}^{n}}\gamma -\alpha ,\]

 

where \[\tan \alpha =\frac{a}{b}\] is the general solution.

 

\[(-\sqrt{{{a}^{2}}+{{b}^{2}}})\le \,a\cos \theta +b\sin \theta \le \,(\sqrt{{{a}^{2}}+{{b}^{2}}})\]

 

The general solution of \[a\cos x+b\sin x=c\] is

 

\[x=2n\pi +{{\tan }^{-1}}\left( \frac{b}{a} \right)\pm {{\cos }^{-1}}\left( \frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)\].

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