Question

How do you solve $\sin x.{\tan ^2}x = \sin x$ ?

Answer 1

In order to solve the above trigonometric equation, rewrite the equation by pulling out common $\sin x$ from both the term and derive the solution by equating every factor one by one equal to zero. Use the fact that the period of function tangent is $\pi$ and the graph of sine function results zero for every $\pi$ interval to obtain the generalised solution of the equation.

Complete step by step solution:
We are given a trigonometric equation $\sin x{\tan ^2}x = \sin x$ .
$\sin x.{\tan ^2}x = \sin x$
Rewriting the equation by transposing $\sin x$ from the right-hand side towards left-hand side of the equation with the help of rules transposing of terms, we get
$\sin x{\tan ^2}x - \sin x = 0$
As we can see $\sin x$ is common in both the terms, so pull out common $\sin x$ from both the terms
$\sin x\left( {{{\tan }^2}x - 1} \right) = 0$ -----(1)
One by one we will divide the above equation with $\sin x$ and later with ${\tan ^2}x - 1$
So, first Dividing both sides of the equation(1) with $\sin x$
$\dfrac{{\sin x}}{{\sin x}}\left( {{{\tan }^2}x - 1} \right) = 0 \times \dfrac{1}{{\sin x}} \\\ {\tan ^2}x - 1 = 0 \;$
Simplifying it further, we get
${\tan ^2}x = 1$
Since $\tan \left( {\dfrac{\pi }{4}} \right) = 1$ , so writing this in above equation we have
${\tan ^2}x = \tan \left( {\dfrac{\pi }{4}} \right) \\\ \tan x = \pm \tan \left( {\dfrac{\pi }{4}} \right) \;$
The period of function tangent is $\pi$ as the graph of tangent repeats itself after every $\pi$ interval. Generalising the solution we get
$x = n\pi \pm \dfrac{\pi }{4}$ where n is an integer--------(2)
Now dividing both sides of the equation (1) with ${\tan ^2}x - 1$ , we get

$$\dfrac{{\sin x}}{{\left( {{{\tan }^2}x - 1} \right)}}\left( {{{\tan }^2}x - 1} \right) = 0 \times \dfrac{1}{{\left( {{{\tan }^2}x - 1} \right)}} \\\ \sin x = 0 \\\ x = {\sin ^{ - 1}}\left( 0 \right) \;$$

Since the graph sine function results zero for after every $\pi$ interval
$x = n\pi$--------(3)
From equation (2) and (3) we can conclude the solution as
$x = n\pi ,n\pi \pm \dfrac{\pi }{4}$
Therefore, the solution to the given trigonometric equation is $x = n\pi \,or\,x = n\pi \pm \dfrac{\pi }{4}$
So, the correct answer is “$x = n\pi \,or\,x = n\pi \pm \dfrac{\pi }{4}$”.

Note : Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$ for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$ for all x in its domain.
We know that $\sin ( - \theta ) = - \sin \theta \cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta$
Therefore, $\sin \theta$ and $\tan \theta$ and their reciprocals, $\cos ec\theta$ and $\cot \theta$ are odd functions whereas $\cos \theta$ and its reciprocal $\sec \theta$ are even functions.
Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.