Question

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

Answer 1

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.