# Question 84672

Dec 6, 2017

$x = k \pi , k \in \mathbb{Z}$

#### Explanation:

$\cos 3 x \cdot \cos x - 1 = 0$

From tables take the formula for cos(3x):

$\left(4 C o {s}^{3} x - 3 \cos x\right) \cdot \cos x - 1 = 0$

$4 C o {s}^{4} x - 3 {\cos}^{2} x - 1 = 0$

This can be transformed in a second degree equation where $y = {\cos}^{2} \left(x\right)$:

$4 {y}^{2} - 3 y - 1$

$y = \frac{3 \pm \sqrt{{3}^{2} - 4 \cdot 4 \cdot \left(- 1\right)}}{2 \cdot 4}$

$y = \frac{3 \pm \sqrt{9 + 16}}{8} = \frac{3 \pm \sqrt{25}}{8}$

$y = \frac{3 \pm 5}{8}$

$y = \frac{8}{8} = 1 \mathmr{and} y = - \frac{2}{8} = - \frac{1}{4}$

Now replace y by Cox^2x:

${\cos}^{2} x = 1 \mathmr{and} {\cos}^{2} x = - \frac{1}{4}$

cosx=+-1 or impossible

$x = k \pi$

Dec 6, 2017

$x = k \pi$

#### Explanation:

Use trig identity:
$\cos a . \cos b = \left(\frac{1}{2}\right) \left[\cos \left(a + b\right) + \cos \left(a - b\right)\right]$
In this case:
$\cos 3 x . \cos x = \frac{\cos 4 x + \cos 2 x}{2} - 1 = 0$
cos 4x + cos 2x - 2 = 0
Replace in the equation cos 4x by $\left(2 {\cos}^{2} 2 x - 1\right)$ -->
$2 {\cos}^{2} 2 x + \cos 2 x - 3 = 0$
Solve this quadratic equation for cos 2x.
Since a + b + c = 0, use shortcut. The 2 real roots are:
$\cos 2 x = 1$, and $\cos 2 x = \frac{c}{a} = - \frac{3}{2}$ (rejected)
cos 2x = 1. Unit circle -->
$2 x = 0 + 2 k \pi$ --> $x = k \pi$
$2 x = 2 \pi + 2 k \pi$ --> x = pi + kpi General answer x = kpi Check. x = pi$- \to$cos 3x = cos pi = - 1$- \to$cos x = cos pi = -1. cos 3x.cos x - 1 = (-1)(-1) - 1 = 0#. Proved