Question

How do you find the general solutions for $2{{\cos }^{2}}4x-1=0$?

Answer 1

Hint: We have been given a quadratic equation of $\cos x$. We divide both sides of the equation by the constant 2. Then we take the square root on both sides of the equation. From that we find the exact and general solutions for the equation $2{{\cos }^{2}}4x-1=0$ for $x$.

Complete step by step answer:
The given equation of $\cos x$ is $2{{\cos }^{2}}4x-1=0$. Simplifying we get $2{{\cos }^{2}}4x=1$.
We divide both sides of the equation by the constant 2 and get
$\begin{align}
  & \dfrac{2{{\cos }^{2}}4x}{2}=\dfrac{1}{2} \\
 & \Rightarrow {{\cos }^{2}}4x=\dfrac{1}{2} \\
\end{align}$
We take square root on both sides of the equation. As the equation is a quadratic one, the number of roots will be 2 and they are equal in value but opposite in sign.
$\begin{align}
  & \sqrt{{{\cos }^{2}}4x}=\sqrt{\dfrac{1}{2}} \\
 & \Rightarrow \cos 4x=\pm \dfrac{1}{\sqrt{2}} \\
\end{align}$
We know that in the principal domain or the periodic value of $0\le x\le \pi $ for $\cos x$, if we get $\cos a=\cos b$ where $0\le a,b\le \pi $ then $a=b$.
We have $\cos 4x=\dfrac{1}{\sqrt{2}}$, the value of \[\cos \left( \dfrac{\pi }{4} \right)\] as $\dfrac{1}{\sqrt{2}}$ in the domain of $0\le x\le \pi $.
We have $\cos 4x=-\dfrac{1}{\sqrt{2}}$, the value of \[\cos \left( \dfrac{3\pi }{4} \right)\] as $-\dfrac{1}{\sqrt{2}}$ in the domain of $0\le x\le \pi $.
Therefore, $\cos 4x=\pm \dfrac{1}{\sqrt{2}}$ gives $x=\dfrac{\pi }{4},\dfrac{3\pi }{4}$ as primary value.
The general solution for this will be $4x=\left( n\pi \pm \dfrac{\pi }{4} \right)\cup \left( n\pi \pm \dfrac{3\pi }{4} \right)$.
Simplifying the equation, we get $x=\left( \dfrac{n\pi }{4}\pm \dfrac{\pi }{16} \right)\cup \left( \dfrac{n\pi }{4}\pm \dfrac{3\pi }{16} \right)$.

Note: Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty $. In that case we have to use the formula $x=n\pi \pm a$ for $\cos \left( x \right)=\cos a$ where $0\le x\le \pi $. For our given problem $\cos 4x=\pm \dfrac{1}{\sqrt{2}}$, the primary solution is $x=\dfrac{\pi }{4},\dfrac{3\pi }{4}$.
The general solution will be $x=\left( \dfrac{n\pi }{4}\pm \dfrac{\pi }{16} \right)\cup \left( \dfrac{n\pi }{4}\pm \dfrac{3\pi }{16} \right)$. Here $n\in \mathbb{Z}$.