# Solve the equation cos3t=cos2t?

May 22, 2017

$\text{The Soln. Set is } \left\{\frac{2}{5} k \pi : k \in \mathbb{Z}\right\} .$

#### Explanation:

$\cos 2 \theta = \cos 3 \theta .$

$\Rightarrow \cos 3 \theta - \cos 2 \theta = 0.$

$\Rightarrow - 2 \sin \left(\frac{3 \theta + 2 \theta}{2}\right) \cdot \sin \left(\frac{3 \theta - 2 \theta}{2}\right) = 0.$

$\Rightarrow - 2 \sin \left(\frac{5 \theta}{2}\right) \sin \left(\frac{\theta}{2}\right) = 0.$

$\Rightarrow \sin \left(\frac{5 \theta}{2}\right) = 0 , \mathmr{and} , \sin \left(\frac{\theta}{2}\right) = 0.$

Knowing that, $\sin x = o \Rightarrow x = k \pi , k \in \mathbb{Z} ,$ we have,

$\frac{5}{2} \theta = k \pi , \mathmr{and} , \frac{\theta}{2} = k \pi , k \in \mathbb{Z} .$

$\theta = \frac{2}{5} k \pi , \mathmr{and} , \theta = 2 k \pi , k \in \mathbb{Z} .$

$\Rightarrow \theta \in \left\{\frac{2}{5} k \pi\right\} \cup \left\{2 k \pi\right\} , k \in \mathbb{Z} .$

But, we observe that, $\left\{2 k \pi : k \in \mathbb{Z}\right\} \subset \left\{\frac{2}{5} k \pi : k \in \mathbb{Z}\right\} ,$ hence

$\text{The Soln. Set is } \left\{\frac{2}{5} k \pi : k \in \mathbb{Z}\right\} .$

Enjoy Maths.!

May 22, 2017

Solution is $\theta = \frac{2 n \pi}{5}$, where $n$ is an integer.

#### Explanation:

We will use here $\cos B - \cos A = 2 \sin \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$

Hence, we ccan write $\cos 2 \theta = \cos 3 \theta$ as

$\cos 2 \theta - \cos 3 \theta = 0$

or $2 \sin \left(\frac{3 \theta + 2 \theta}{2}\right) \sin \left(\frac{3 \theta - 2 \theta}{2}\right) = 0$

or $2 \sin \left(\frac{5 \theta}{2}\right) \sin \left(\frac{\theta}{2}\right) = 0$

i.e. either $\frac{5 \theta}{2} = n \pi$ or $\frac{\theta}{2} = n \pi$, where $n$ is an integer.

i.e. either $\theta = \frac{2 n \pi}{5}$ or $\theta = 2 n \pi$, where $n$ is an integer.

Observe that the latter set (i.e. where $\theta = 2 n \pi$) is a subset of former set (i.e. where $\theta = \frac{2 n \pi}{5}$). This is so when $n$ is a multiple of $5$.

Hence solution is $\theta = \frac{2 n \pi}{5}$, where $n$ is an integer.

May 22, 2017

$t = 2 k \pi$
$t = \frac{2 k \pi}{5}$

#### Explanation:

cos 3t = cos 2t
$3 t = \pm 2 t$

a. 3t = 2t
t = 0 or $t = 2 k \pi$

b. 3t = - 2t
5t = 0 or $5 t = 2 k \pi$
$t = \frac{2 k \pi}{5}$
Check by calculator, with k = 1
$t = \frac{2 \pi}{5} = {72}^{\circ}$ --> $\cos 2 t = \cos {144}^{\circ} = - 0.809$ -->
$\cos 3 t = \cos {216}^{\circ} = - 0.809$. OK
$t = 2 \pi$ --> $\cos 2 t = \cos 4 \pi = 1$ and $\cos 3 t = \cos 6 \pi = 1$. OK