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

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

Explanation

Solution

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(0,2π)x \in (0,2\pi ) . Here it is open intervals. So we don’t include 00 and 2π2\pi for the value of ‘x’.

Complete step-by-step answer :
Given, sinxtanxsinx=0\sin x\tan x - \sin x = 0
Taking sine function common we have,
sinx(tanx1)=0\sin x(\tan x - 1) = 0
Now using the zero product principle we have
sinx=0\sin x = 0 and tanx1=0\tan x - 1 = 0 .
Now take the first factor sinx=0\sin x = 0
We know that sine function is zero at 00 , π\pi and 2π2\pi .
That is sinx=0\sin x = 0 then x=0,π,2πx = 0,\pi ,2\pi .
But x(0,2π)x \in (0,2\pi ) that is an open interval. Hence we have only one value of ‘x’. That is x=πx = \pi .
Now take the second factor
tanx1=0\tan x - 1 = 0
tanx=1\tan x = 1
We know that at x=π4x = \dfrac{\pi }{4} and at x=π+π4x = \pi + \dfrac{\pi }{4} the value of tanx=1\tan x = 1 .
(we know tan(π+θ)=tanθ\tan (\pi + \theta ) = \tan \theta and its lies in third quadrant, tangent is positive in third quadrant)
Thus we have, x=π4x = \dfrac{\pi }{4} and x=5π4x = \dfrac{{5\pi }}{4} . Both x(0,2π)x \in (0,2\pi ) .
(Because we know that the value of 54=1.25\dfrac{5}{4} = 1.25 ).
Thus the solution of sinxtanxsinx=0\sin x\tan x - \sin x = 0 for x(0,2π)x \in (0,2\pi ) is x=πx = \pi , x=π4x = \dfrac{\pi }{4} and x=5π4x = \dfrac{{5\pi }}{4} .
So, the correct answer is “ x=πx = \pi , x=π4x = \dfrac{\pi }{4} and x=5π4x = \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.