Question

How do you find the general solutions for $\sin x = \cos 2x - 1$?

Answer 1

Hint: First transform $\cos 2x$ in terms of $\sin x$. After that move all terms to one side. Then factorize the equation by factorization method to get the solution of $\sin x$. After that, we will get the general solution for the equation.

Complete step-by-step answer:
Given that,
$ \Rightarrow \sin x = \cos 2x - 1$
Before proceeding with the solution we should understand the concept of trigonometric equations. The equations which involve trigonometric functions like sin, cos, tan, cot, sec, etc. are called trigonometric equations.
We already know that the values of sin x and cos x repeat after an interval of $2\pi $ . Also, the values of tan x repeat after an interval of $\pi $ . A general solution is one that involves the integer ‘n’ and gives all solutions of a trigonometric equation. Also, the character ‘Z’ is used to denote the set of integers. Thus, a solution generalized using periodicity is known as the general solution.
We know that,
$\cos 2x = 1 - 2{\sin ^2}x$
Put this in the above equation,
$ \Rightarrow \sin x = 1 - 2{\sin ^2}x - 1$
Simplify the terms,
$ \Rightarrow \sin x = - 2{\sin ^2}x$
Move the term on the left side,
$ \Rightarrow 2{\sin ^2}x + \sin x = 0$
Take $\sin x$ common from the left side,
$ \Rightarrow \sin x\left( {2\sin x + 1} \right) = 0$
For $\sin x = 0$,
As the value of \[\sin x\] gives 0 on $0,\pi ,2\pi , \ldots $
So, the general solution is $n\pi $.
For $2\sin x + 1 = 0$,
Move constant part on the right side,
$ \Rightarrow 2\sin x = - 1$
Divide both sides by 2,
$ \Rightarrow \sin x = - \dfrac{1}{2}$
As we know, the value of $\sin x$ is negative in the third and fourth quadrant.
For the third quadrant,
$ \Rightarrow \sin x = \sin \left( {\pi + \dfrac{\pi }{6}} \right)$
So, the general solution will be,
$ \Rightarrow x = 2n\pi + \pi + \dfrac{\pi }{6}$
Simplify the terms,
\[ \Rightarrow x = \left( {2n + 1} \right)\pi + \dfrac{\pi }{6}\]
For the fourth quadrant,
$ \Rightarrow \sin x = \sin \left( { - \dfrac{\pi }{6}} \right)$
So, the general solution will be,
$ \Rightarrow x = 2n\pi - \dfrac{\pi }{6}$

Hence, the general solution of $\sin x = \cos 2x - 1$ is $n\pi ,\left( {2n + 1} \right)\pi + \dfrac{\pi }{6}$ and $2n\pi - \dfrac{\pi }{6}$

Note:
In solving trigonometric equations we need to remember the formulas, the standard values of angles, and the identities because then it becomes easy. We in a hurry can make a mistake in applying the cofunction identities as we can write cos in place of sin and sin in place of cos while solving for the general solution.