A) \[\frac{1}{2}\]
B) 2
C) \[\frac{11}{2}\]
D) \[\frac{46}{5}\]
Correct Answer: D
Solution :
Given \[2\cos \theta +\sin \theta =1\] Squaring both sides, we get \[{{(2\cos \theta +\sin \theta )}^{2}}={{1}^{2}}\] \[\Rightarrow \]\[4{{\cos }^{2}}+{{\sin }^{2}}\theta +4\sin \theta \cos \theta =1\] \[\Rightarrow \]\[3{{\cos }^{2}}\theta +(co{{s}^{2}}\theta +si{{n}^{2}}\theta )+4sin\theta \cos \theta =1\] \[\Rightarrow \]\[3{{\cos }^{2}}\theta +\cancel{1}+4\sin \theta \cos \theta =\cancel{1}\] \[\Rightarrow \]\[3{{\cos }^{2}}\theta +4\sin \theta \cos \theta =0\] \[\Rightarrow \]\[\cos \theta (3cos\theta +4sin\theta )=0\] \[\Rightarrow \]\[\cos \theta (3cos\theta +4sin\theta )=0\] \[\Rightarrow \]\[3\cos \theta -4\sin \theta \] \[\Rightarrow \]\[\frac{-3}{4}=\tan \theta =\sqrt{{{\sec }^{2}}\theta -1}=\frac{-3}{4}\] \[\left( \because \tan \theta =\sqrt{{{\sec }^{2}}\theta -1} \right)\] \[\Rightarrow \]\[{{\sec }^{2}}\theta -1={{\left( \frac{-3}{4} \right)}^{2}}=\frac{9}{16}\] \[\Rightarrow \]\[{{\sec }^{2}}\theta =\frac{9}{16}+1=\frac{25}{16}\Rightarrow \sec \theta =\frac{5}{4}\] or\[\] ...(1) Now, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Rightarrow {{\sin }^{2}}\theta +{{\left( \frac{4}{5} \right)}^{2}}=1\] \[{{\sin }^{2}}\theta +\frac{4}{5}=1\Rightarrow {{\sin }^{2}}\theta =1-\frac{16}{25}=\frac{9}{25}\] \[\sin \theta =\pm \frac{3}{5}\] ?(2) Taking\[\left( \sin \theta =+\frac{3}{5} \right)\]because\[\left( \sin \theta =-\frac{3}{5} \right)\]cannot satisfy the given equation. Therefore; \[7\cos \theta +6\sin \theta \] \[=7\times \frac{4}{5}+6\times \frac{3}{5}=\frac{28}{5}+\frac{18}{5}=\frac{46}{5}\]
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