# 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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