A) \[x=n\pi +{{\cos }^{-1}}\left( \frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)\]
B) \[x=2n\pi -{{\tan }^{-1}}\left( \frac{b}{a} \right)\]
C) \[x=2n\pi -{{\tan }^{-1}}\left( \frac{b}{a} \right)\pm {{\cos }^{-1}}\left( \frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)\]
D) \[x=2n\pi +{{\tan }^{-1}}\left( \frac{b}{a} \right)\pm {{\cos }^{-1}}\left( \frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)\]
Correct Answer: D
Solution :
\[\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\cos x+\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\sin x=\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\] \[\Rightarrow \] \[\cos \left( x-{{\cos }^{-1}}\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)=\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\] \[\Rightarrow \] \[x-{{\cos }^{-1}}\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}={{\cos }^{-1}}\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\] General solution is, \[x-{{\cos }^{-1}}\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=2n\pi \pm {{\cos }^{-1}}\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\] or \[x=2n\pi \pm {{\cos }^{-1}}\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}+{{\cos }^{-1}}\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\] x = \[2n\pi +{{\tan }^{-1}}\frac{b}{a}\pm {{\cos }^{-1}}\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]. Trick: Put\[a=b=c=1\], then \[\cos \left( x-\frac{\pi }{4} \right)\]=\[\cos \frac{\pi }{4}\] \[\Rightarrow \] \[x=2n\pi +\frac{\pi }{4}\pm \frac{\pi }{4}\] which is given by option (d).
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