Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Recently I worked on a problem where I had to solve $$\cos(3x)+\cos(x)=0$$
When I tried calculating it by evaluating $x$, $x=2nπ+\dfrac{\pi}{4}$, $x=2n\pi-\dfrac{\pi}{4}$, or $x=2n\pi+\dfrac{\pi}{2}$ was the solution I reached.
Unfortunately, it was apparently wrong. Is there another way to solve this problem?
So, we have ${\cos(2x + x) + \cos(x) = 0}$. Using the compound trig formulae:
$${\Rightarrow \cos(2x)\cos(x) - \sin(2x)\sin(x) + \cos(x) = 0}$$
Using the double angle identity for ${\sin(2x)}$, and ${\cos(2x)}$ gives
$${(2\cos^2(x) - 1)\cos(x) - 2\cos(x)\sin^2(x) + \cos(x)= 0}$$
$${\Rightarrow 2\cos^3(x) - 2\cos(x)\left(1 - \cos^2(x)\right)=0}$$
$${\Rightarrow 4\cos^3(x) - 2\cos(x) = 0}$$
$${\Rightarrow \cos(x)\left(4\cos^2(x) - 2\right)=0}$$
So either ${\cos(x)=0}$ or ${4\cos^2(x) - 2 = 0}$. Can you take it from here?
I think it is easier to use addition formula $\cos(a)+\cos(b)=2\cos(\frac{a+b}2)\cos(\frac{a-b}2)$ in this case rather than expanding $\cos(nx)$.
You get directly the $\frac \pi 2+k\pi$ and $\frac \pi 4+\frac k2\pi\ $ lines from the product:
$$2\cos(x)\cos(2x)=0$$
$$\cos(3x)+ \cos(x)=0$$ $$\cos(3x)=- \cos(x)$$ $$\cos(3x)= \cos(\pi-x)$$ $$3x= 2k\pi\pm (\pi-x)$$ $$x=\frac{(2k+1)\pi}{4}, \ \ \frac{(2k-1)\pi}{2}$$ $$x\in\{\frac{(2k+1)\pi}{4}\}\cup\{\frac{(2k-1)\pi}{2}\}$$ Where, $k$ is an integer i.e. $k=0, \pm1, \pm2, \ldots$
Use the cosine of sums formula to decompose $\cos 3x$.
$$\cos 3x = \cos (2x + x) = \cos 2x \cdot \cos x - \sin x \cdot \sin 2x$$
The identities $\cos 2x = 2 \cos^2 x - 1$ and $\sin 2x = 2 \sin x \cos x$ are widely known. These are also derived from the cosine of sums and sine of sums formulas respectively. Plugging them into our last result, we have
\begin{align} \cos 3x &= \left(2 \cos^2 x - 1\right) \cos x - \sin x \cdot 2 \sin x \cos x \\ &= 2\cos^3 x - \cos x - 2\sin^2 x \cdot \cos x \\ \end{align}
Since $\sin^2 x = 1 - \cos^2 x$, we have
\begin{align} \cos 3x &= 2\cos^3 x - \cos x - 2\sin^2 x \cdot \cos x \\ &= 2\cos^3 x - \cos x - 2 \left(1 - \cos^2 x\right) \cos x \\ &= 2\cos^3 x - \cos x - 2 \cos x + 2 \cos^3 x \\ &= 4\cos^3 x - 3 \cos x \end{align}
At this point, we have eliminated the $\sin$'s and $\cos 3x$ is expressed entirely in terms of a polynomial of $\cos x$. Plugging this into our equation $\cos^3 x + \cos x = 0$, we have
\begin{align} \left(4\cos^3 x - 3 \cos x\right) + \cos x &= 0 \\ 4\cos^3 x - 2 \cos x &= 0 \\ 2\cos^3 x - \cos x &= 0 \end{align}
Taking $y = \cos x$, this reduces to the polynomial
$$2y^3 - y = 0 \Rightarrow y(2y^2 - 1) = 0 \Rightarrow y(y - \sqrt{2}/2)(y + \sqrt{2}/2) = 0$$
which gives three distict roots $y = 0$, $y = -\sqrt{2}/2$ and $y = \sqrt{2}/2$. Then, we plug-in $y = \cos x$ into each and solve them separately.
For $y = 0$, we have $\cos x = 0$. This holds for $x = \pi/2$ plus any integer multiple of $\pi$. These are exactly the points where $\cos x$ intersects the $x$-axis.
For $y = -\sqrt{2}/2$, we have $\cos x = -\sqrt{2}/2$. From here, we have exactly the bottom points of the cosine curve $\pm \pi/4$.
For $y = +\sqrt{2}/2$, we have $\cos x = +\sqrt{2}/2$. From here, we have exactly the peak points of the cosine curve $\pm \pi/4$.
Taking the union of these solution sets, we get the solution set for $\cos^3 x + \cos x = 0$ which are exactly the multiples of $\pi/4$ that are not multiples of $\pi/2$.
Yet another method using
$cos(3x)=4cos^3(x)-3cos(x)$
$cos(3x)-cos(x)=0$
$4cos^3(x)=4cos(x)$
$cos^3(x)=cos(x)$
$cos(x)=0$ ----(1)
Or
$cos^2(x)=1$
$sin^2(x)=0$ -----(2)
Use (1) and (2) to find general solutions
Symmetry is always a powerful and nice instrument $$ \eqalign{ & 0 = \cos (3x) + \cos x = \cr & = \cos (2x + x) + \cos (2x - x) = \cr & = \cos (2x)\cos x - \sin (2x)\sin x + \cos (2x)\cos x + \sin (2x)\sin x = \cr & = 2\cos (2x)\cos x \cr} $$
and then $$ \eqalign{ & \cos (2x) = 0\quad \vee \quad \cos x = 0 \cr & 2x = {\pi \over 2} + k\pi \quad \vee \quad x = {\pi \over 2} + j\pi \cr & x = {\pi \over 4} + k{\pi \over 2}\quad \vee \quad x = {\pi \over 2} + j\pi \cr & x = {\pi \over 4} + 2k{\pi \over 4}\quad \vee \quad x = 2{\pi \over 4} + 4j{\pi \over 4} \cr & x = \left( {2k + 1} \right){\pi \over 4}\quad \vee \quad \left( {2j + 1} \right){\pi \over 2}\quad \left| {\,k,j \in Z} \right. \cr} $$
