Question

If \[5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9,\] then the value of \[\cos 4x\] is:
a) \[ - \dfrac{3}{5}\]
b) \[ \dfrac{1}{3}\]
c) \[ \dfrac{2}{9}\]
d) \[ - \dfrac{7}{9}\]

Answer 1

Hint: \[{\tan ^2}x\] can be written in terms of \[{\sec ^2}x\] and \[{\sec ^2}x\] can be also written as \[{\sec ^2}x.\cos 4x\] is nothing but \[\left[ {\cos 2\left( {2x} \right)} \right].\] They to find out the answer by simplifying the concept. One can write the trigonometric values into its equivalent values and assume any one value to be to ease the calculation.
For Ex:\[\cos 2x = 2{\cos ^2}x - 1\] and \[\cos 4x = 2{\cos ^2}2x - 1\]and further, we can write \[\cos 4x = 2{\left( {\cos 2x} \right)^2} - 1\].
Thus, \[\cos 4x = 2{\left( {2{{\cos }^2}x - 1} \right)^2} - 1\] and can be further simplified.

Complete step-by-step solution:
Given: We known that \[{\tan ^2}x = {\sec ^2}x - 1\] and \[\cos 2x = 2{\cos ^2}x - 1,\] remember these are standard values that we use and not according to our wish.
Therefore, \[5\left( {{{\sec }^2}x - 1 - {{\cos }^2}x} \right) = 2\left[ {2{{\cos }^2}x - 1} \right] + 9\]
Let’s
Take \[{\cos ^2}x = t\]
\[
   \Rightarrow 5\left( {\dfrac{1}{t} - 5 - 5t} \right) = 2\left( {2t - 1} \right) + 9 \\
   \Rightarrow \dfrac{5}{t} - 5 - 5t = 4t + 7 \\
   \Rightarrow 5 - 5t - 5{t^2} = 4{t^2} + 7t \\
 \]
\[
   \Rightarrow 9{t^2} + 15t - 3t - 5 = 0 \\
   \Rightarrow \left( {3t + 5} \right)\left( {3t - 1} \right) = 0 \\
   \Rightarrow t = - \dfrac{3}{5}\,\,\,or\,\,t = \dfrac{1}{3} \\
 \]
Now the main tricky part of the solution comes here, as we know \[t = {\cos ^2}x\] and the mange of \[{\cos ^2}x\] is [0,1]. Therefore, is cannot be a negative value.
Therefore,
\[{\cos ^2}x\]is not equal to \[ - \dfrac{3}{5}\]
Thus, we are left with
\[{\cos ^2}x = \dfrac{1}{3}\]
Now, we know that \[\cos 4x\] can be written in terms of \[\cos 2x\] i.e. \[\cos 4x\] is also equal to \[\left( {\cos 2x\left( {2x} \right)} \right)\]
Thus,
\[
  \cos 2\left( {2x} \right) = 2{\cos ^2}2x - 1 \\
   = 2{\left[ {2{{\cos }^2}x - 1} \right]^2} - 1 \\
 \]
\[ = 2{\left[ {2 \times \dfrac{1}{3} - 1} \right]^2} - 1\][Putting the value of \[{\cos ^2}x\] we get]
\[
  \, = \dfrac{2}{9} - 1 \\
   = \dfrac{7}{9} \\
 \]
Thus, the value of \[\cos 4x\] is \[ - \dfrac{7}{9}\] .
Therefore, according to our question option (d) is correct.

Note: In these types of questions students often make mistakes by putting the wrong formula. Keep this in mind while solving also do not get misled up with your concepts like writing \[\cos 4x\] as \[\cos 3x + \cos x\] it is completely correct.