Question

How do you solve the inequality \[\sin 3x < \sin x\]?

Answer 1

Hint:For any inequality when you are solving you should know the sign of inequality changes when you multiply minus sign both the side, and rest solution can be done same as that for equals sign is done, nor any other assumption should be needed.

Formulae Used: Basic trigonometric identities are used here,
\[\sin 3x = 3\sin x - 4{\sin ^3}x\], \[1 - {\sin ^2}x = {\cos ^2}x\], \[{\cos ^2}x - {\sin ^2}x = \cos 2x\].

Complete step by step solution:
 In the given equation \[\sin 3x < \sin x\]
Firstly either you can transfer right hand part into left side or left hand part into right side, here let us transfer the right hand part into left side, we get:

\[\sin 3x - \sin x < 0\](for every positive value of “sinx”)

Let's expand the terms and solve further we get:
\[
\sin 3x - \sin x < 0 \\
(3\sin x - 4{\sin ^3}x) - \sin x < 0\,(\sin 3x = 3\sin x - 4{\sin ^3}x) \\
2\sin x - 4{\sin ^3}x < 0 \\
2\sin x(1 - 2{\sin ^2}x) < 0 \\
2\sin x(1 - {\sin ^2}x - {\sin ^2}x) < 0 \\
2\sin x({\cos ^2}x - {\sin ^2}x) < 0\,\,(1 - {\sin ^2}x = {\cos ^2}x) \\
2\sin x(\cos 2x) < 0\,\,({\cos ^2}x - {\sin ^2}x = \cos 2x) \\
\]

This is our required solution for the given inequality.

Additional Information: If you are given the equals to sign with inequalities then also the process would be same the only change would be in describing the range of the quantity, that is closed bracket would be used instead of an open bracket. Here range is not defined because it was not provided in question.

Note: Inequality basically defines the region of the quantity whosoever for which the inequality is used for, that is it gives you a range of possible values for the quantity you are finding for. In this range real as also complex range also occurs. Here you have to be sure of the positive values of “sinx” because if you take negative values also then signs of inequality will change accordingly.