# Question 0f638

Dec 2, 2017

See below.

#### Explanation:

This equation is equivalent to

$\sqrt{{\sin}^{2} x} = \sqrt{{\cos}^{2} \left(3 x\right)}$

now squaring both sides

${\sin}^{2} x = {\cos}^{2} \left(3 x\right) \Rightarrow \left(\sin x + \cos 3 x\right) \left(\sin x - \cos 3 x\right) = 0$

and now knowing that

$\cos 3 x = {\cos}^{3} x - 3 \cos x {\sin}^{2} x$

$\left\{\begin{matrix}\sin x + {\cos}^{3} x - 3 \cos x {\sin}^{2} x = 0 \\ \sin x - {\cos}^{3} x + 3 \cos x {\sin}^{2} x = 0\end{matrix}\right.$

Those equations to be completely solved require the solution of a cubic polynomial. We left this as an exercise.

Dec 2, 2017

A different method to the one below.

#### Explanation:

We have $\left\mid \sin x \right\mid = \left\mid \cos 3 x \right\mid$. But by definition, $\cos 3 x = \sin \left(\frac{\pi}{2} - 3 x\right)$.

So the equation becomes $\left\mid \sin x \right\mid = \left\mid \sin \left(\frac{\pi}{2} - 3 x\right) \right\mid$.

This equation is true iff $\sin x = \sin \left(\frac{\pi}{2} - 3 x\right)$ or $\sin x = - \sin \left(\frac{\pi}{2} - 3 x\right) = \sin \left(3 x - \frac{\pi}{2}\right)$

Finding general solutions will be left as an exercise

Dec 5, 2017

$\left(x = \frac{\pi}{8} + \frac{k \pi}{4}\right)$

#### Explanation:

$| \sin x | = | \cos 3 x |$

From the tables we know that :

$\textcolor{b l u e}{\cos 3 x = 4 {\cos}^{3} x - 3 \cos x}$,

$| \sin x | = | 4 {\cos}^{3} x - 3 \cos x |$

$\pm \sin x = \left(4 {\cos}^{2} x - 3\right) \cos x$

$\pm \tan x = \left(4 {\cos}^{2} x - 3\right)$

$\textcolor{b l u e}{1 + {\tan}^{2} x = \frac{1}{\cos} ^ 2 x}$

${\tan}^{2} x = {\left(\left(4 {\cos}^{2} x - 3\right)\right)}^{2} = \left(16 {\cos}^{4} x - 24 {\cos}^{2} x + 9\right)$

$\frac{1}{\cos} ^ 2 x = 16 {\cos}^{4} x - 24 {\cos}^{2} x + 9 + 1$

$1 = 16 {\cos}^{6} x - 24 {\cos}^{4} x + 10 C o {s}^{2} x$

I could solve this equation with $y = {\cos}^{2} x$

$16 {y}^{3} - 24 {y}^{2} + 10 y - 1 = 0$

but it is beyond my capabilities.

If we play a little with the trigonometric circle we can find

that $x = \pm \frac{\pi}{4} + k \pi$ is one of the solutions, as

$| \sin \left(x\right) | = | \sin \left(\frac{\pi}{4}\right) + k \pi | = \frac{\sqrt{2}}{2}$

and

$| \cos \left(3 x\right) | = | \cos \left(3 \frac{\pi}{4}\right) + 3 k \pi | = \frac{\sqrt{2}}{2}$

color(red)(x=pi/4+kpi/2

so $x = \frac{\pi}{4} \to \cos \left(x\right) = \frac{\sqrt{2}}{2} \to y = 0.5$

We just need to divide the third degree equation by (y-0.5) and obtain the remainder solutions:

$16 {y}^{2} - 16 y + 2 = 0$

$8 {y}^{2} - 8 y + 1 = 0$

$y = \frac{8 \pm \sqrt{{8}^{2} - 4 \cdot 8 \cdot 1}}{2 \cdot 8}$

$y = \frac{8 \pm \sqrt{64 - 32}}{16}$

$y = \frac{8 \pm \sqrt{32}}{16}$

$y = \frac{8 \pm 4 \sqrt{2}}{16}$

$y = \frac{2 \pm \sqrt{2}}{4}$

$\to {\cos}^{2} x = y$

cosx=sqrt((2+-sqrt(2))/(4)

$\cos x = \frac{\sqrt{2 \pm \sqrt{2}}}{2}$

$\cos x = \cos \left(\pm \frac{\pi}{8}\right) \mathmr{and} \cos x = \cos \left(\pm \frac{3 \pi}{8}\right)$

$x = \pm \frac{\pi}{8} + 2 k \pi \mathmr{and} x \pm 3 \frac{\pi}{8} + 2 k \pi$

$\textcolor{red}{x = \frac{\pi}{8} + k \frac{\pi}{2}}$

Remember the first solution:

color(red)(x=pi/4+kpi/2

and gives

$\textcolor{g r e e n}{x = \frac{\pi}{8} + \frac{k \pi}{4}}$

Dec 5, 2017

x = pi/8 + (kpi)/4
x = - pi/4 + kpi
x = - (3pi/4) + kpi

#### Explanation:

Note that sin x = cos (pi/2 - x)
$I \cos \left(\frac{\pi}{2} - x\right) I = I \cos 3 x I$
We have to solve these equations:
$\cos \left(\frac{\pi}{2} - x\right) = \cos 3 x$, and
$\cos \left(\frac{\pi}{2} - x\right) = - \cos 3 x$
A. $\cos \left(\frac{\pi}{2} - x\right) = \cos 3 x$
$\left(\frac{\pi}{2} - x\right) = \pm 3 x$
a. $\frac{\pi}{2} - x = 3 x$ --> $4 x = \frac{\pi}{2} + 2 k \pi$ --> $x = \frac{\pi}{8} + \frac{k \pi}{4}$
b. $\frac{\pi}{2} - x = - 3 x$ --> $2 x = - \frac{\pi}{2} + 2 k \pi$ --> $x = - \frac{\pi}{4} + k \pi$
B. $\cos \left(\frac{\pi}{2} - x\right) = - \cos 3 x = \cos \left(3 x + \pi\right)$
$\left(\frac{\pi}{2} - x\right) = \pm \left(3 x + \pi\right)$
a. $\frac{\pi}{2} - x = 3 x + \pi$ --> $4 x = - \frac{\pi}{2} + 2 k \pi$ -->
$x = - \frac{\pi}{4} + k \pi$
b. $\frac{\pi}{2} - x = - 3 x - \pi$ --> $2 x = - \frac{3 \pi}{2} + 2 k \pi$ -->
$x = - \left(\frac{3 \pi}{4}\right) + k \pi$
Finally, the general solutions are:
x = pi/8 + (kpi)/4; x = - pi/4 + kpi; x = -((3pi)/4) + kpi#
Check.
$x = \frac{\pi}{8} = {22}^{\circ} 5$ --> $\sin 22.5 = 0.38$ -->
$\cos 67.5 = 0.38$. Proved
$x = - 45$ --> $\sin \left(- 45\right) = - \frac{\sqrt{2}}{2}$ -->
$\cos \left(- 135\right) = - \frac{\sqrt{2}}{2}$. Proved
$x = - \frac{3 \pi}{4}$ --> $\sin x = - \frac{\sqrt{2}}{2}$ -->
$\cos \left(\frac{9 \pi}{4}\right) = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$. Proved.
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