# How to find the exact solutions which lie in [0, 2π) for cos(5x) = −cos(2x)?

May 12, 2018

(i)x=pi/7,(3pi)/7,(5pi)/7,(7pi)/7,(9pi)/7,(11pi)/7,(13pi)/7.tox in [0,2pi]
$\left(i i\right) x = \frac{\pi}{3} , \frac{3 \pi}{3} , \frac{5 \pi}{3.} \to x \in \left[0 , 2 \pi\right]$

#### Explanation:

We know that,

color(violet)((I)cosC+cosD=2cos((C+D)/2)cos((C-D)/2)

Here,

$\cos 5 x = - \cos 2 x$

=>color(violet)(cos5x+cos2x=0...toApply(I)

=>color(violet)(2cos((5x+2x)/2)cos((5x-2x)/2)=0

$\implies \cos \left(\frac{7 x}{2}\right) \cos \left(\frac{3 x}{2}\right) = 0$

=>color(blue)(cos((7x)/2)=0 or cos((3x)/2)=0

color(blue)((i)cos((7x)/2)=0

=>(7x)/2=pi/2,(3pi)/2,(5pi)/2,(7pi)/2,(9pi)/2,(11pi)/2,(13pi)/2, (15pi)/2,...

$\implies 7 x = \pi , 3 \pi , 5 \pi , 7 \pi , 9 \pi , 11 \pi , 13 \pi , 15 \pi , \ldots$

=>color(red)(x=pi/7,(3pi)/7,(5pi)/7,(7pi)/7,(9pi)/7,(11pi)/7,(13pi)/7.tox in [0,2pi]

color(blue)((ii)cos((3x)/2)=0

$\implies \frac{3 x}{2} = \frac{\pi}{2} , \frac{3 \pi}{2} , \frac{5 \pi}{2} , \frac{7 \pi}{2} , \frac{9 \pi}{2} , \ldots$

$\implies 3 x = \pi , 3 \pi , 5 \pi , 7 \pi , 9 \pi , \ldots$

=>color(red)(x=pi/3,(3pi)/3,(5pi)/3.to x in [0,2pi]

Note:

The general solution is:

$\implies \frac{7 x}{2} = \left(2 k + 1\right) \frac{\pi}{2} , k \in \mathbb{Z} \mathmr{and} \frac{3 x}{2} = \left(2 k + 1\right) \frac{\pi}{2} , k \in \mathbb{Z}$

$\implies x = \left(2 k + 1\right) \frac{\pi}{7} , k \in Z \mathmr{and} x = \left(2 k + 1\right) \frac{\pi}{3} , k \in Z$

May 13, 2018

pi/3 ; pi ; (5pi)/3
pi/7 ; (3pi)/7 ; (5pi)/7 ; pi; (9pi)/7 ; (11pi)/7 ; (13pi)/7

#### Explanation:

cos 5x = - cos 2x
$\cos 5 x = \cos \left(2 x + \pi\right)$
Unit circle and property of cos x -->
$5 x = \pm \left(2 x + \pi\right) + 2 k \pi$
a. $5 x = 2 x + \pi + 2 k \pi = 2 x + \left(2 k + 1\right) \pi$
$3 x = \left(2 k + 1\right) \pi$
$x = \left(2 k + 1\right) \frac{\pi}{3}$
k = 0 --> $x = \frac{\pi}{3}$ ; k = 1 --> $x = \pi$ ; k = 2 --> $x = \frac{5 \pi}{3}$
b. $5 x = - 2 x - \pi + 2 k \pi = - 2 x + \left(2 k - 1\right) \pi$
$7 x = \left(2 k - 1\right) \pi$
$x = \left(2 k - 1\right) \frac{\pi}{7}$
k = 0 --> $x = - \frac{\pi}{7}$ or $x = \frac{13 \pi}{7}$ (co- terminal)
k = 1 --> $x = \frac{\pi}{7}$ ; k = 2 --> $x = \frac{3 \pi}{7}$ ; k = 3 -->
$x = \frac{5 \pi}{7}$ ; k = 4 --> $x = \pi$ ; k = 5 --> $x = \frac{9 \pi}{7}$ ;
k = 6 --> $x = \frac{11 \pi}{7}$