# Question fcb2e

Jun 12, 2017

$x \approx 0.251$

#### Explanation:

$\cos x \cos 3 x = 0.707$

$f \left(x\right) = \cos x \cos 3 x - 0.707 = 0$

Use Newton method to find a zero of $f$

${x}_{n + 1} = {x}_{n} - \frac{f \left({x}_{n}\right)}{f ' \left({x}_{n}\right)}$

$f ' \left(x\right) = - 2 \sin 2 x - 2 \sin 4 x$

$0.707 = \cos \left(\frac{1}{4} \pi\right)$, so a good ${x}_{0}$ would be $\frac{\frac{1}{4} \pi}{3} = 0.262$

x_1=0.262-(cos(0.262)cos(3(0.262))-0.707)/(2sin(2(0.262)-2sin(4(0.262))=0.253#

${x}_{2} = 0.251$

${x}_{3} = 0.251$

${x}_{4} = 0.251$

The iterations approach $0.251$ to $0.251$ is a solution to $\cos x \cos 3 x = 0.707$

Jun 12, 2017

$x = \pm {14}^{\circ} 48$

#### Explanation:

cos x.cos 3x = 0.707
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:
$\cos x . \cos 3 x = \left(\frac{1}{2}\right) \left(\cos 2 x + \cos 4 x\right)$-->
cos 2x + cos 4x = 2(0.707) = 1.414
Call 2x = X,
cos X + cos 2X - 1.414 = 0.
Replace cos 2X by (2cos^2 X - 1):
$\cos X + \left(2 {\cos}^{2} X - 1\right) - 1.414 = 0$
2cos^2 X + cos X - 2.414 = 0.
Solve this quadratic equation for cos X.
$D = {d}^{2} = {b}^{2} - 4 a c = 1 + 19.28 = 20.28$ --> $d = \pm 4.50$
There are 2 real roots:
$\cos X = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{1}{4} \pm \frac{4.50}{4} = - 0.25 \pm 1.125$
a. cos X = - 1.375 (rejected as < - 1)
b. cos 2x = cos X = 0.875
Calculator and unit circle give:
$2 x = \pm {28}^{\circ} 96$ --> $x = \pm {14}^{\circ} 48$
Check by calculator:
$x = {14}^{\circ} 48$ --> $\cos x = 0.97$ --> $3 x = {43}^{\circ} 44$ --> $\cos 3 x = 0.73$.
cos x.cos 3x = 0.97(0.73) = 0.71. Proved.