# How to solve sin3x<sinx?

Jun 13, 2018

The solution is $x \in \left(\frac{\pi}{4} , \frac{3}{4} \pi\right) \cup \left(\pi , \frac{5}{4} \pi\right) \cup \left(\frac{7}{4} \pi , 2 \pi\right)$, $\left[2 \pi\right]$

#### Explanation:

To solve this trigonometric inequality, we need

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

The inequality is

$\sin 3 x < \sin x$

$\implies$, $\sin 3 x - \sin x < 0$

Let $f \left(x\right) = \sin 3 x - \sin x$

The period of $f \left(x\right)$ is $T = 2 \pi$

Study the functio on the interval $I = \left[0 , 2 \pi\right]$

Therefore,

$f \left(x\right) = 3 \sin x - 4 {\sin}^{3} x - \sin x$

$= 2 \sin x - 4 {\sin}^{3} x$

$= 2 \sin x \left(1 - {\sin}^{2} x\right)$

Let $2 \sin x \left(1 - {\sin}^{2} x\right) = 0$

The solutions to this equation are

$\left\{\begin{matrix}\sin x = 0 \\ 1 - 2 {\sin}^{2} x = 0\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = 0 + k \pi \\ x = \frac{\pi}{4} + k \pi \\ x = \frac{5}{4} \pi + k \pi\end{matrix}\right.$

Build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$\frac{\pi}{4}$$\textcolor{w h i t e}{a a a a}$$\frac{3}{4} \pi$$\textcolor{w h i t e}{a a a a}$$\pi$$\textcolor{w h i t e}{a a a a}$$\frac{5}{4} \pi$$\textcolor{w h i t e}{a a a a}$$\frac{7}{4} \pi$$\textcolor{w h i t e}{a a a a}$$2 \pi$

$\textcolor{w h i t e}{a a a a}$$\sin x$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$

$\textcolor{w h i t e}{a}$$1 - 2 {\sin}^{2} x$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$

Therefore,

$f \left(x\right) < 0$ when $x \in \left(\frac{\pi}{4} , \frac{3}{4} \pi\right) \cup \left(\pi , \frac{5}{4} \pi\right) \cup \left(\frac{7}{4} \pi , 2 \pi\right)$, $\left[2 \pi\right]$

graph{sin(3x)-sinx [-8.54, 37.08, -9.16, 13.64]}

Jun 13, 2018

sin 3x - sin x < 0
f(x) = 2cos (2x). sin x < 0
First solve the 2 basic trig equations to find the end-points (critical points)
1. g(x) = cos 2x = 0 --> $2 x = \frac{\pi}{2}$, and $2 x = \frac{3 \pi}{2}$ -->
x = pi/4; x = (3pi)/4; x = (5pi)/4, and x = (7pi)/4
2. h(x) = sin x = 0 --> x = 0; x = pi; and x = 2pi
Next, to algebraically solve the trig inequality, create a sign chart that shows the variation of the 2 functions g(x) and h(x) when x varies from 0 to $2 \pi$ through the critical points.
$\left(0 , \frac{\pi}{4} , \frac{3 \pi}{4} , \pi , \frac{5 \pi}{4} , \frac{7 \pi}{4} , 2 \pi\right)$
The sign (+ or -) of the function f(x) is the resulting sign of the 2 function g(x) and h(x)
The solutions are the intervals
$\left(\frac{\pi}{4} , \frac{3 \pi}{4}\right)$ where g(x) < 0, and h(x) > 0 and --> f(x) < 0
$\left(\pi , \frac{5 \pi}{4}\right)$ where g(x) > 0, and h(x) < 0, and --> f(x) < 0
$\left(\frac{7 \pi}{4} , 2 \pi\right)$ where g(x) > 0, and h(x) < 0 , and --> f(x) < 0
You also can solve the trig inequality by using graphing calculator.
The parts of the graph that lie below the x-axis represent the answers inside the interval $\left(0 , 2 \pi\right)$
Note. The graph shows x in radians. Exp. x = pi = 3.14;
x = pi/2 = 1.57