Question

Find the general solution of the equation $2\cos 2x=3.2{{\cos }^{2}}x-4$.

Answer 1

Here we are going to simplify the given equation using Trigonometric formulas and convert it into some simple formats like $\sin x=\sin y$ or $\cos x=\cos y$ or $\tan x=\tan y$ and then we can find the solution of the equation as $x=y$.

Complete step by step answer:
Given that,
$\begin{aligned} & 2\cos 2x=3.2{{\cos }^{2}}x-4 \\\ & 2\cos 2x=3\left( 2{{\cos }^{2}}x \right)-4........\left( \text{i} \right) \end{aligned}$
We know that, $\cos \left( \text{A}+\text{B} \right)=\cos \text{A}\text{.}\cos \text{B}-\sin \text{A}\sin \text{B}$ hence
$\begin{aligned} & \cos \left( \text{A}+\text{A} \right)=\cos \text{A}.\cos \text{A}-\sin \text{A}.\sin \text{A} \\\ & \text{cos2A}={{\cos }^{2}}\text{A}-{{\sin }^{2}}\text{A} \end{aligned}$
We have the trigonometry identity as ${{\sin }^{2}}\text{A}+{{\cos }^{2}}\text{A}=1$ then the above equation modified as
$\begin{aligned} & \cos 2\text{A}={{\cos }^{2}}\text{A}-\left( 1-{{\cos }^{2}}\text{A} \right) \\\ & \cos 2\text{A}=2{{\cos }^{2}}\text{A}-1 \\\ \end{aligned}$
From the above formula we are substituting $2{{\cos }^{2}}x=1+\cos 2x$ in equation $\left( \text{i} \right)$, we have
$\begin{aligned} & 2\cos 2x=3\left( 1+\cos 2x \right)-4 \\\ & 2\cos 2x=3+3\cos 2x-4 \\\ & 1=\cos 2x........\left( \text{ii} \right) \end{aligned}$
We know that the values of $\cos x$ are varies as shown in the below figure

From the above figure we can say that for values like $2\pi ,4\pi ,6\pi ,...$ we have $\cos x=1$ then from the equation $\left( \text{ii} \right)$ we can write
$\begin{aligned} & \cos 2n\pi =\cos 2x \\\ & 2x=2n\pi \\\ & x=n\pi ,n\in I \end{aligned}$

Note:
While using the formula $\cos 2x=2{{\cos }^{2}}x-1$ substitute the value of $2{{\cos }^{2}}x$ but not substitute the value of $\cos 2x$ why because if you substitute the value of $\cos 2x$ then the equation turns into polynomial equation and the we get $2$ values for the solution of $x$