Question

How do you solve $\sin x - \tan x = 0$?

Answer 1

Hint:A general solution is one which gives all solutions of a given trigonometric equation where n is integer and $n \in Z$.
The above equation can be easily solved by putting the values of the given trigonometric function.
In this question, $\tan x$ can be written in the form of $\sin x$ and $\cos x$.

Complete step by step answer:
We know, $\tan x = \dfrac{{\sin x}}{{\cos x}}$
Putting the value of $\tan x$in $\sin x - \tan x = 0$
$ \Rightarrow \sin x - \dfrac{{\sin x}}{{\cos x}} = 0$
Taking LCM and solving the above equation,
$ \Rightarrow \dfrac{{\sin x\cos x - \sin x}}{{\cos x}} = 0$
By cross multiplication,
$ \Rightarrow \sin x\cos x - \sin x = 0 \cdot \cos x$
$ \Rightarrow \sin x\cos x - \sin x = 0$
Take $\sin x$ common,
$ \Rightarrow \sin x(\cos x - 1) = 0$
A factor should be zero.
$ \Rightarrow \sin x = 0$ or $\cos x - 1 = 0$
$ \Rightarrow \sin x = 0$ or $\cos x = 1$
We know, the general solution of $\sin x = 0$ is $x = n\pi $
and for $\cos x = 1$
We know, $\cos {0^ \circ } = 1$
So, $\cos x = \cos {0^ \circ }$
$ \Rightarrow \cos x = \cos \alpha $
$ \Rightarrow x = 2n\pi \pm \alpha $
Here, $\alpha = {0^ \circ }$
So, $x = 2n\pi \pm 0$
$ \Rightarrow x = 2n\pi $
Thus, $x = n\pi $ or $x = 2n\pi $ , where $n \in Z$ and n is integer.
Additional information:
You can also solve this by taking $\tan x$ to the right-hand side.
$ \Rightarrow \sin x = \tan x$
$ \Rightarrow \sin x = \dfrac{{\sin x}}{{\cos x}}$
$ \Rightarrow \sin x\cos x = \sin x$
Bring $\sin x$ to the left-hand side,
$ \Rightarrow \sin x\cos x - \sin x = 0$
Take $\sin x$ common
$ \Rightarrow \sin x(\cos x - 1) = 0$
Factors should be zero
$ \Rightarrow \sin x = 0$ or $\cos x - 1 = 0$
$ \Rightarrow \sin x = 0$ or $\cos x = 1$
We know, $\cos {0^ \circ } = 1$
So, $\cos x = \cos {0^ \circ }$
$ \Rightarrow \cos x = \cos \alpha $
$ \Rightarrow x = 2n\pi \pm \alpha $
Here, $\alpha = {0^ \circ }$
So, $x = 2n\pi \pm 0$
$ \Rightarrow x = 2n\pi $

Thus, $x = n\pi $ or $x = 2n\pi $ $n \in Z$ and n is integer.

Note: The general solution of $\sin x = 0$ is $x = n\pi $ , general solution of $\cos x = 1$ is $x = 2n\pi $ , general solution of $\cos x = 0$ is $x = (2n + 1)\dfrac{\pi }{2}$ and general solution of $\tan x = 0$ is $x = n\pi $ where $n \in Z$ and n is integer.
Range of $\sin x = [ - 1,1]$ ,domain is all real numbers and period $ = 2\pi $
Range of $\cos x = [ - 1,1]$ ,domain is all real numbers and period $ = 2\pi $
Range of $\tan x = $ all real numbers, domain is $R - (2k + 1)\dfrac{\pi }{2}$ , $k \in Z$, k is integer and R is real number , period $ = \pi $
Always try to reduce the trigonometric equations in simpler functions like $\sin \theta $, $\cos \theta $ to make it easier to solve.
Always bring the right-hand side values to the left-hand side so as to factorize and make the factor zero for each.
Signs of all trigonometric functions should be taken care of for every interval or quadrant.