Quantitative Aptitude Ques 1410

Question: If $ 2\cos \theta -\sin \theta =\frac{1}{\sqrt{2}}(0{}^\circ <\theta <90{}^\circ ), $ then the value of $ 2\sin \theta +\cos \theta $ is

Options:

A) $ \frac{1}{\sqrt{2}} $

B) $ \sqrt{2} $

C) $ \frac{3}{\sqrt{2}} $

D) $ \frac{\sqrt{2}}{3} $

Show Answer

Answer:

Correct Answer: C

Solution:

  • Given, $ 2\cos \theta -\sin \theta =\frac{1}{\sqrt{2}} $ … (i) Let $ 2\sin \theta +\cos \theta =x $ … (ii) On squaring Eqs. (i) and (ii) and then adding, we get $ 4{{\cos }^{2}}\theta +{{\sin }^{2}}-4\sin \theta \cdot \cos \theta +4{{\sin }^{2}}\theta $ $ +{{\cos }^{2}}\theta +4\sin \theta \cos \theta =\frac{1}{2}+x^{2} $

$ \Rightarrow $ $ 4({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )+({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )=\frac{1}{2}+x^{2} $

$ \Rightarrow $ $ 4+1=\frac{1}{2}+x^{2} $ $ [\because {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1] $

$ \Rightarrow $ $ x^{2}=5-\frac{1}{2}=\frac{9}{2} $

$ \therefore $ $ x=\frac{3}{\sqrt{2}} $

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