Question

How do I solve \[\sin x\tan x - \sin x = 0\] for \[(0,2\pi )\] ?

Answer 1

Hint: We can take sine as common, since sine is present in both the terms. Using the zero product principle we get two factors. After solving the factors we can get the value of ‘x’. In the problem they are given that \[x \in (0,2\pi )\] . Here it is open intervals. So we don’t include \[0\] and \[2\pi \] for the value of ‘x’.

Complete step-by-step answer:
Given, \[\sin x\tan x - \sin x = 0\]
Taking sine function common we have,
 \[\sin x(\tan x - 1) = 0\]
Now using the zero product principle we have
 \[\sin x = 0\] and \[\tan x - 1 = 0\] .
Now take the first factor \[\sin x = 0\]
We know that sine function is zero at \[0\] , \[\pi \] and \[2\pi \] .
That is \[\sin x = 0\] then \[x = 0,\pi ,2\pi \] .
But \[x \in (0,2\pi )\] that is an open interval. Hence we have only one value of ‘x’. That is \[x = \pi \] .
Now take the second factor
 \[\tan x - 1 = 0\]
 \[\tan x = 1\]
We know that at \[x = \dfrac{\pi }{4}\] and at \[x = \pi + \dfrac{\pi }{4}\] the value of \[\tan x = 1\] .
(we know \[\tan (\pi + \theta ) = \tan \theta \] and its lies in third quadrant, tangent is positive in third quadrant)
Thus we have, \[x = \dfrac{\pi }{4}\] and \[x = \dfrac{{5\pi }}{4}\] . Both \[x \in (0,2\pi )\] .
(Because we know that the value of \[\dfrac{5}{4} = 1.25\] ).
Thus the solution of \[\sin x\tan x - \sin x = 0\] for \[x \in (0,2\pi )\] is \[x = \pi \] , \[x = \dfrac{\pi }{4}\] and \[x = \dfrac{{5\pi }}{4}\] .
So, the correct answer is “ \[x = \pi \] , \[x = \dfrac{\pi }{4}\] and \[x = \dfrac{{5\pi }}{4}\] ”.

Note: Remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant. Also sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively.