Solve the equation cos^2x+cos^2(3x)=1?

Mar 28, 2017

$x = \left(k \pi \pm \frac{\pi}{4}\right) \cup \left(\frac{k \pi}{2} \pm \frac{\pi}{8}\right)$ where $k$ is an integer.

Explanation:

${\cos}^{2} \left(3 x\right) = 1 - {\cos}^{2} x = {\sin}^{2} x$ so

$\left(\cos \left(3 x\right) + \sin \left(x\right)\right) \left(\cos \left(3 x\right) - \sin \left(x\right)\right) = 0$

or

1) $\cos \left(3 x\right) = - \sin \left(x\right) = \sin \left(- x\right) = \cos \left(\frac{\pi}{2} - \left(- x\right)\right)$

then

$3 x = 2 k \pi \pm \left(\frac{\pi}{2} + x\right)$

i.e. either $2 x = 2 k \pi + \frac{\pi}{2}$ and $x = k \pi + \frac{\pi}{4}$

or $4 x = 2 k \pi - \frac{\pi}{2}$ and $x = \frac{k \pi}{2} - \frac{\pi}{8}$

2) $\cos \left(3 x\right) - \sin \left(x\right) = 0$

or

$\cos \left(3 x\right) = \sin \left(x\right) = \cos \left(\frac{\pi}{2} - x\right)$

so

$3 x = 2 k \pi \pm \left(\frac{\pi}{2} - x\right)$

i.e. either $4 x = 2 k \pi + \frac{\pi}{2}$ and $x = \frac{k \pi}{2} + \frac{\pi}{8}$

or $2 x = 2 k \pi - \frac{\pi}{2}$ and $x = k \pi - \frac{\pi}{4}$

so the solutions are

$x = \left(k \pi \pm \frac{\pi}{4}\right) \cup \left(\frac{k \pi}{2} \pm \frac{\pi}{8}\right)$ where $k$ is an integer.

Mar 28, 2017

$\left\{x : x = \left.\left(\frac{\pi}{4} + n \pi\right) , \left(\frac{\text{-"pi)/8+(npi)/2),(pi/8+(npi)/2),(("-} \pi}{4} + n \pi\right)\right. , n \in \mathbb{Z}\right\}$ or equivalently $\left\{x : x = \frac{n \pi}{8} , n \ne 4 m , n \in \mathbb{Z} , m \in \mathbb{Z}\right\}$

Explanation:

${\cos}^{2} x + {\cos}^{2} \left(3 x\right) = 1$

Subtract ${\cos}^{2} x$ from both sides
${\cos}^{2} \left(3 x\right) = 1 - {\cos}^{2} x$

Using the Pythagorean Identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$,
$1 - {\cos}^{2} x = {\sin}^{2} x$, therefore
${\cos}^{2} \left(3 x\right) = {\sin}^{2} x$

Subtract ${\sin}^{2} x$ from both sides
${\cos}^{2} \left(3 x\right) - {\sin}^{2} x = 0$

Factor the left side using the difference of squares
$\left(\cos \left(3 x\right) + \sin x\right) \left(\cos \left(3 x\right) - \sin x\right) = 0$

Time to find the value of $x$

Case 1: $\cos \left(3 x\right) + \sin x = 0$

Subtract $\sin x$ from both sides
$\cos \left(3 x\right) = \text{-} \sin x$

Since "-"sintheta=sin("-"theta),
"-"sinx=sin("-"x), therefore
$\cos \left(3 x\right) = \sin \left(\text{-} x\right)$

Since $\sin \theta = \cos \left(\frac{\pi}{2} - \theta\right)$ and $\cos \left(\theta\right) = \cos \left(\theta + 2 n \pi\right) , n \in \mathbb{Z}$,
$\sin \left(\text{-} x\right) = \cos \left(\frac{\pi}{2} + x + 2 n \pi\right)$, therefore
$\cos \left(3 x\right) = \cos \left(\frac{\pi}{2} + x + 2 n \pi\right)$

Since $\cos \theta = \cos \left(\text{-} \theta\right)$,
$3 x = \frac{\pi}{2} + x + 2 n \pi \mathmr{and} 3 x = \frac{\text{-} \pi}{2} - x - 2 n \pi$

Case 1.1: $3 x = \frac{\pi}{2} + x + 2 n \pi$

Subtract $x$ from both sides
$2 x = \frac{\pi}{2} + 2 n \pi$

Divide both sides by $2$
$\textcolor{b l u e}{x = \frac{\pi}{4} + n \pi , n \in \mathbb{Z}}$

Case 1.2: $3 x = \frac{\text{-} \pi}{2} - x - 2 n \pi$

Add $x$ to both sides
$4 x = \frac{\text{-} \pi}{2} - 2 n \pi$

Divide both sides by $4$
$\textcolor{b l u e}{x = \frac{\text{-} \pi}{8} - \frac{n \pi}{2} , n \in \mathbb{Z}}$

Case 2: $\cos \left(3 x\right) - \sin x = 0$

Add $\sin x$ to both sides
$\cos \left(3 x\right) = \sin x$

Since $\sin \theta = \cos \left(\frac{\pi}{2} - \theta\right)$ and $\cos \left(\theta\right) = \cos \left(\theta + 2 n \pi\right) , n \in \mathbb{Z}$,
$\sin x = \cos \left(\frac{\pi}{2} - x + 2 n \pi\right)$, therefore
$\cos \left(3 x\right) = \cos \left(\frac{\pi}{2} - x + 2 n \pi\right)$

Since $\cos \theta = \cos \left(\text{-} \theta\right)$,
$3 x = \frac{\pi}{2} - x + 2 n \pi \mathmr{and} 3 x = \frac{\text{-} \pi}{2} + x - 2 n \pi$

Case 2.1: $3 x = \frac{\pi}{2} - x + 2 n \pi$

Add $x$ to both sides
$4 x = \frac{\pi}{2} + 2 n \pi$

Divide both sides by $4$
$\textcolor{b l u e}{x = \frac{\pi}{8} + \frac{n \pi}{2} , n \in \mathbb{Z}}$

Case 2.2: $3 x = \frac{\text{-} \pi}{2} + x - 2 n \pi$

Subtract $x$ from both sides
$2 x = \frac{\text{-} \pi}{2} - 2 n \pi$

Divide both sides by $2$
$\textcolor{b l u e}{x = \frac{\text{-} \pi}{4} - n \pi , n \in \mathbb{Z}}$

Since $n$ can be any integer, any term with $\text{-} n$ can be replaced with $n$, so compiling everything gives
$\textcolor{red}{\left\{x : x = \left.\left(\frac{\pi}{4} + n \pi\right) , \left(\frac{\text{-"pi)/8+(npi)/2),(pi/8+(npi)/2),(("-} \pi}{4} + n \pi\right)\right. , n \in \mathbb{Z}\right\}}$ or equivalently $\textcolor{red}{\left\{x : x = \frac{n \pi}{8} , n \ne 4 m , n \in \mathbb{Z} , m \in \mathbb{Z}\right\}}$