Question

The general solution of the equation, \[2\cos 2x=3.2{{\cos }^{2}}x-4\]is: -
(a) \[x=2n\pi ,n\in I\]
(b) \[x=n\pi ,n\in I\]
(c) \[x=\dfrac{n\pi }{4},n\in I\]
(d) \[x=\dfrac{n\pi }{2},n\in I\]

Answer 1

Hint: Convert into its half angle by using the identity, \[\cos 2x=2{{\cos }^{2}}x-1\]. Form a quadratic equation in \[\cos x\] and solve this equation to get the values of \[\cos x\]. If any value does not lie in the range of \[\cos x\], i.e. [-1, 1] then eliminate that value. For the trigonometric equation as: - \[x=n\pi \pm a,n\in I\].

Complete step-by-step solution
We have been provided with the equation: -
\[\Rightarrow \] \[2\cos 2x=3.2{{\cos }^{2}}x-4\].
First of all, note that in R.H.S the coefficient of \[{{\cos }^{2}}x\] is 3.2. It is not a decimal number but it denotes 3 multiplied by 2.
So, the equation becomes,
\[\Rightarrow 2\cos 2x=6{{\cos }^{2}}x-4\]
Using the half angle formula to write \[\cos 2x\] in terms of \[{{\cos }^{2}}x\], we have,
\[\cos 2x=2{{\cos }^{2}}x-1\], hence the required equation becomes,
\[\begin{align}
  & \Rightarrow 2\left( 2{{\cos }^{2}}x-1 \right)=6{{\cos }^{2}}x-4 \\
 & \Rightarrow 4{{\cos }^{2}}x-2=6{{\cos }^{2}}x-4 \\
 & \Rightarrow 2{{\cos }^{2}}x=2 \\
 & \Rightarrow {{\cos }^{2}}x=1 \\
\end{align}\]
Now, we know that, \[{{\cos }^{2}}0=1\], therefore,
\[\Rightarrow {{\cos }^{2}}x={{\cos }^{2}}0\]
The general solution for trigonometric equations of the form: - \[{{\cos }^{2}}x={{\cos }^{2}}a\] is given by \[x=n\pi \pm a,n\in I\].
\[\begin{align}
  & \Rightarrow {{\cos }^{2}}x={{\cos }^{2}}0 \\
 & \Rightarrow x=n\pi \pm 0 \\
 & \Rightarrow x=n\pi ,n\in I \\
\end{align}\]
Hence, option (B) is the correct answer.

Note: One may note that in the starting we have written 3.2 as 3 multiplied by 2. This is because in higher mathematics dot (.) is the symbol for multiplication. In the above question, we have converted \[\cos 2x\] and written it in term of \[{{\cos }^{2}}x\], but one can also convert \[{{\cos }^{2}}x\] in the R.H.S into \[\cos 2x\] by using the formula: - \[{{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}\]. So, at last we will have a equation of the form, \[\cos 2x=\cos a\] whose general solution is given as: - \[2x=2n\pi +a,n\in I\].