| 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 [SSC (10+2) 2012] |
A) \[\frac{1}{\sqrt{2}}\]
B) \[\sqrt{2}\]
C) \[\frac{3}{\sqrt{2}}\]
D) \[\frac{\sqrt{2}}{3}\]
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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