Trigonometric Equations Question 334
Question: The general solution of $ a\cos x+b\sin x=c, $ where $ a,b,c $ are constants
Options:
A) $ x=n\pi +{{\cos }^{-1}}( \frac{c}{\sqrt{a^{2}+b^{2}}} ) $
B) $ x=2n\pi -{{\tan }^{-1}}( \frac{b}{a} ) $
C) $ x=2n\pi -{{\tan }^{-1}}( \frac{b}{a} )\pm {{\cos }^{-1}}( \frac{c}{\sqrt{a^{2}+b^{2}}} ) $
D) $ x=2n\pi +{{\tan }^{-1}}( \frac{b}{a} )\pm {{\cos }^{-1}}( \frac{c}{\sqrt{a^{2}+b^{2}}} ) $
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Answer:
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 ( x-{{\cos }^{-1}}\frac{a}{\sqrt{a^{2}+b^{2}}} )=\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 ( x-\frac{\pi }{4} ) $ = $ \cos \frac{\pi }{4} $
$ \Rightarrow $ $ x=2n\pi +\frac{\pi }{4}\pm \frac{\pi }{4} $ which is given by option (d).
BETA
