Question

How do you solve the identity $\cos 2x+\cos 4x=0$?

Answer 1

Hint: We have been given a trigonometric equation. We convert it into a quadratic equation of $\cos 2x$. We assume the value of $\cos 2x$ as the variable $m$. Then we use quadratic solving to solve the problem. We use the quadratic formula to solve the value of the $m$. We have the solution in the form of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for general equation of $a{{x}^{2}}+bx+c=0$.

Complete step-by-step solution:
We know the multiple angle formula for ratio cos where $\cos 2x=2{{\cos }^{2}}x-1$.
So, $\cos 4x=2{{\cos }^{2}}2x-1$. We replaced the value of $x$ with $2x$.
The given equation of $\cos 2x+\cos 4x=0$ becomes $\cos 2x+2{{\cos }^{2}}2x-1=0$. We assume the term $\cos 2x$ as the variable $m$.
The revised form of the equation is
\[\begin{align}
  & \cos 2x+2{{\left( \cos 2x \right)}^{2}}-1=0 \\
 & \Rightarrow 2{{m}^{2}}+m-1=0 \\
\end{align}\].
We solve the quadratic equation with formula of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for general equation of $a{{x}^{2}}+bx+c=0$. In this case of \[2{{m}^{2}}+m-1=0\], we will get root values for $m$.
We get $m=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times 2\times \left( -1 \right)}}{2\times 2}=\dfrac{-1\pm \sqrt{9}}{4}=\dfrac{-1\pm 3}{4}=-1,\dfrac{1}{2}$
So, values of $m$ are $m=-1,\dfrac{1}{2}$. This gives $\cos 2x=-1,\dfrac{1}{2}$.
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 $\cos 2x=-1$, the value of $\cos \left( \pi \right)$ as $-1$. $0\le \pi \le \pi $.
We have $\cos 2x=\dfrac{1}{2}$, the value of $\cos \left( \dfrac{\pi }{3} \right)$ as $\dfrac{1}{2}$. $0<\dfrac{\pi }{3}<\pi $.
Therefore, $\cos 2x=-1,\dfrac{1}{2}$ gives $2x=\pi ,\dfrac{\pi }{3}$ as primary value.
The general solution will be $2x=\left( n\pi \pm \pi \right)\cup \left( n\pi \pm \dfrac{\pi }{3} \right)$. Here $n\in \mathbb{Z}$.
Simplifying the solution, we get the solution for $x$ as \[x=\left( \dfrac{n\pi \pm \pi }{2} \right)\cup \left( \dfrac{n\pi }{2}\pm \dfrac{\pi }{6} \right)\].

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 $\cos a=\cos b$ where $0\le a,b\le \pi $. For our given problem $\cos 2x=-1,\dfrac{1}{2}$, the primary solution is $x=\dfrac{\pi }{2},\dfrac{\pi }{6}$.