Question 248 of 298MediumMCQ
If 5(tan2x−cos2x)=2cos2x+9, then the value of cos4x is:
Solutions · 1
Official solution
Given that,
\[5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9\]
\[ \Rightarrow 5\left( {{{{{\sin }^2}x} \over {{{\cos }^2}x}} - {{\cos }^2}x} \right) = 2\left( {2{{\cos }^2}x - 1} \right) + 9\]
Let \[{\cos ^2}x = t,\] then we have
\[5\left( {{{1 - t} \over t} - t} \right) = 2\left( {2t - 1} \right) + 9\]
\[ \Rightarrow 5\left( {{{1 - t - {t^2}} \over t}} \right) = 4t - 2 + 9\]
\[ \Rightarrow 5 - 5t - 5{t^2} = 4{t^2} + 7t\]
\[ \Rightarrow 9{t^2} + 12t - 5 = 0\]
\[ \Rightarrow 9{t^2} + 15t - 3t - 5 = 0\]
\[ \Rightarrow 3t\left( {3t + 5} \right) - 1\left( {3t + 5} \right) = 0\]
\[ \Rightarrow \left( {3t + 5} \right)\left( {3t - 1} \right) = 0\]
\[\therefore\] \[t = {1 \over 3}\] and \[t = - {5 \over 3}\]
If \[t = - {5 \over 3}\] then \[{\cos ^2}x\] is negative
So , t can not be \[ - {5 \over 3}\].
So, correct value of \[t = {1 \over 3}\] then \[\cos {}^2x = t = {1 \over 3}\]
\[\therefore\,\,\,\] \[\cos 4x\]
\[ = 2{\cos ^2}2x - 1\]
\[ = 2{\left[ {2{{\cos }^2}x - 1} \right]^2} - 1\]
\[ = 2.{\left[ {2.{1 \over 3} - 1} \right]^2} - 1\]
\[ = 2.{\left( { - {1 \over 3}} \right)^2} - 1\]
\[ = {2 \over 9} - 1\]
\[ = - {7 \over 9}\]
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