Question

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

Answer 1

Hint: 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.