Question

The general solution of the equation, \[2\cos 2x=3.2{{\cos }^{2}}x-4\] .
A. \[2n\pi ,n\in I\]
B. \[n\pi ,n\in I\]
C. \[\dfrac{n\pi }{4},n\in I\]
D. \[\dfrac{n\pi }{2},n\in I\]

Answer 1

Hint: In the question, we have to find the general solution of \[2\cos 2x=3.2{{\cos }^{2}}x-4\] . We know the formula that, \[\cos 2x=2{{\cos }^{2}}x-1\] . Using this formula, replace \[2{{\cos }^{2}}x\] by \[\cos 2x+1\] . Now, find the value of \[\cos 2x\] . We get \[\cos 2x=1\] . We can write it as, \[\cos 2x=\cos 2n\pi \] .The general solution of \[\cos 2x=1\] is \[x=n\pi \] .

Complete step-by-step answer:
According to the question, we have an equation,
\[2\cos 2x=3.2{{\cos }^{2}}x-4\] …………….(1)
We know the formula, \[\cos 2x=2{{\cos }^{2}}x-1\] …………………(2)
On solving equation (2), we get
\[\cos 2x=2{{\cos }^{2}}x-1\]
\[\Rightarrow \cos 2x-1=2{{\cos }^{2}}x\] …………………..(3)
From equation (1) and equation (3), we get
\[2\cos 2x=3.2{{\cos }^{2}}x-4\]
\[\Rightarrow 2\cos 2x=3(1+\cos 2x)-4\]
Solving it further, we get
\[\begin{align}
  & \Rightarrow 2\cos 2x=3+3\cos 2x-4 \\
 & \Rightarrow 2\cos 2x=3\cos 2x-1 \\
 & \Rightarrow 1=3\cos 2x-2\cos 2x \\
\end{align}\]
\[\Rightarrow 1=\cos 2x\] …………………(4)
We know that \[\cos 0=1\] . Now using this in equation (4), we get
\[\begin{align}
  & \cos 2x=1 \\
 & \Rightarrow \cos 2x=\cos 0 \\
 & \Rightarrow 2x=0 \\
\end{align}\]
We also know that \[\cos 2\pi =1\] . Now using this in equation (4), we get
\[\begin{align}
  & \cos 2x=1 \\
 & \Rightarrow \cos 2x=\cos 2\pi \\
 & \Rightarrow 2x=2\pi \\
\end{align}\]
We also know that, \[\cos 4\pi =1\] . Now using this in equation (4), we get
\[\begin{align}
  & \cos 2x=1 \\
 & \Rightarrow \cos 2x=\cos 4\pi \\
 & \Rightarrow 2x=4\pi \\
\end{align}\]
We can see that x can be 0, \[2\pi \] , \[4\pi \] , \[6\pi \] etc.
We can observe that x is the multiple of \[2\pi \] ……………….(5)
Now, from equation (4) and equation (5), we get
\[\cos 2x=1\]
\[\begin{align}
  & \Rightarrow \cos 2x=\cos 2n\pi \\
 & \Rightarrow 2x=2n\pi \\
 & \Rightarrow x=n\pi \\
\end{align}\]
Hence, option (B) is the correct one.

Note: We can also solve this question by using another method.
We know the formula, \[\cos 2x=2{{\cos }^{2}}x-1\] .
Now putting this in \[2\cos 2x=3.2{{\cos }^{2}}x-4\] .
\[\begin{align}
  & 2\cos 2x=3.2{{\cos }^{2}}x-4 \\
 & \Rightarrow 2(2{{\cos }^{2}}x-1)=6{{\cos }^{2}}x-4 \\
 & \Rightarrow 4{{\cos }^{2}}x-2=6{{\cos }^{2}}x-4 \\
 & \Rightarrow 2({{\cos }^{2}}x-1)=0 \\
 & \Rightarrow (\cos x+1)(\cos x-1)=0 \\
\end{align}\]
 \[\cos x=-1\] ……………(1) or \[\cos x=1\] ……………………(2)
We know that \[\cos 0=1\] . Now using this in equation (1), we get
\[\begin{align}
  & \cos x=1 \\
 & \Rightarrow \cos x=\cos 0 \\
 & \Rightarrow x=0 \\
\end{align}\]
We also know that \[\cos 2\pi =1\] . Now using this in equation (1), we get
\[\begin{align}
  & \cos x=1 \\
 & \Rightarrow \cos x=\cos 2\pi \\
 & \Rightarrow x=2\pi \\
\end{align}\]
We also know that, \[\cos 4\pi =1\] . Now using this in equation (1), we get
\[\begin{align}
  & \cos x=1 \\
 & \Rightarrow \cos x=\cos 4\pi \\
 & \Rightarrow x=4\pi \\
\end{align}\]
We can see that x can be 0, \[2\pi \] , \[4\pi \] , \[6\pi \] etc.……………………(3)
If \[\cos x=1\] , then x is the multiple of \[2\pi \] .
We know that \[\cos \pi =-1\] . Now using this in equation (2), we get
\[\begin{align}
  & \cos x=-1 \\
 & \Rightarrow \cos x=\cos \pi \\
 & \Rightarrow x=\pi \\
\end{align}\]
We also know that \[\cos 3\pi =-1\] . Now using this in equation (2), we get
\[\begin{align}
  & \cos x=-1 \\
 & \Rightarrow \cos x=\cos 3\pi \\
 & \Rightarrow x=3\pi \\
\end{align}\]
We also know that, \[\cos 5\pi =-1\] . Now using this in equation (2), we get
\[\begin{align}
  & \cos x=-1 \\
 & \Rightarrow \cos x=\cos 5\pi \\
 & \Rightarrow x=5\pi \\
\end{align}\]
We can see that x can be \[\pi \] , \[3\pi \] , \[5\pi \] etc. ……………….(4)
We can observe that x is the odd multiple of \[\pi \].
If \[\cos x=-1\] , then x is the odd multiple of \[\pi \].
From equation (3) and (4), we have 0, \[\pi \] , \[2\pi \] , \[3\pi \] , \[4\pi \] , \[5\pi \] , and \[6\pi \] .
Therefore, x is the multiple of \[\pi \] .
Hence, option (B) is the correct one.