Question

How do you solve $ 2\cos x\sin x - \cos x = 0 $ ?

Answer 1

Hint: Try to bring the equation in such a form where each of its parts can be individually equated with $ 0 $ . It can be done with the help of factorization.
Factorization refers to writing a mathematical term as a product of several smaller terms of the same kind. These smaller terms are known as factors. One of the methods of factorization is by taking the common term out from the objects present in the equation.

Complete step-by-step answer:
Given equation is:
(i)
 $ 2\cos x\sin x - \cos x = 0 $
We can clearly see that the term $ \cos x $ is present in both of the objects in the equation. So, by factorization i.e., taking the common term out and writing the rest of the objects in parentheses, we get:
 $ \cos x(2\sin x - 1) = 0 $
(ii)
Now, we have got our equation as a product of two terms i.e., $ \cos x $ and $ (2\sin x - 1) $ . As we know that if the product of two terms is $ 0 $ then either one of the terms is $ 0 $ or both of them are $ 0 $ . Thus, we can write:
 $ \cos x = 0 $
Or,
 $ 2\sin x - 1 = 0 $
(iii)
While solving $ \cos x = 0 $ , we need to find such values of $ x $ for which $ \cos x $ becomes $ 0 $ .
Since, no interval is given in the question, we will find the general solution for $ x $ .
As we know that $ \cos \dfrac{\pi }{2} = 0 $ and $ \cos \dfrac{{3\pi }}{2} = 0 $ , we notice that the odd multiples of $ \dfrac{\pi }{2} $ provide us with value $ 0 $ for cosine function.
Now, for obtaining an odd multiplier, we need to take a term which is always odd i.e., which comes just after even.
Since, $ 2n $ is a term which will always be even provided the value of $ n $ will be $ 0,1,2,3,.....,\infty $ , its consecutive term i.e., $ 2n + 1 $ will always be odd for the same values of $ n $ .
Therefore, general solution for $ \cos x = 0 $ will be $ x = (2n + 1)\dfrac{\pi }{2} $ where $ n \in Z $
(iv)
Similarly, while solving $ 2\sin x - 1 = 0 $ , shift everything on RHS except $ \sin x $ , it becomes:
 $
  2\sin x = 1 \\
  \sin x = \dfrac{1}{2} \;
  $
Now, we need to find such values of $ x $ for which $ \sin x $ becomes $ \dfrac{1}{2} $
Again, we’ll find the general solution for $ x $ as no interval is given.
So, as we know that $ \sin \dfrac{\pi }{6} = \dfrac{1}{2} $ and the graph of sine function repeats itself for every interval of $ 2\pi $ , we can write that
 $ \sin x = \dfrac{1}{2} $ for $ x = \dfrac{\pi }{6},\dfrac{\pi }{6} + 2\pi ,\dfrac{\pi }{6} + 4\pi ,.......,\dfrac{\pi }{6} + 2n\pi $ where $ n = 0,1,2,3,....,\infty $
Therefore,
  $ \sin x = \dfrac{1}{2} $ for $ x = 2n\pi + \dfrac{\pi }{6} $ where $ n \in Z $ -(eq.1)
Also, we know the identity, $ \sin (\pi - \theta ) = \sin \theta $
Here, putting $ \theta $ as $ \dfrac{\pi }{6} $ , we will get
 $
  \sin (\pi - \dfrac{\pi }{6}) = \sin \dfrac{\pi }{6} \\
  \sin \dfrac{{5\pi }}{6} = \dfrac{1}{2} \;
  $
Again, it will repeat the values after the interval of $ 2\pi $ which gives us,
 $ \sin x = \dfrac{1}{2} $ for $ x = (\pi - \dfrac{\pi }{6}),(\pi - \dfrac{\pi }{6} + 2\pi ),(\pi - \dfrac{\pi }{6} + 4\pi ),.....,(\pi - \dfrac{\pi }{6} + 2n\pi ) $
i.e.,
  $ \sin x = \dfrac{1}{2} $ for $ x = (\pi - \dfrac{\pi }{6}),(3\pi - \dfrac{\pi }{6}),(5\pi - \dfrac{\pi }{6}),.....,((2n + 1)\pi - \dfrac{\pi }{6}) $ where $ n = 0,1,2,3,....,\infty $
Therefore,
 $ \sin x = \dfrac{1}{2} $ for $ x = (2n + 1)\pi - \dfrac{\pi }{6} $ where $ n \in Z $ -(eq.2)
After observing eq.1 and eq.2, we see that the coefficient of $ \dfrac{\pi }{6} $ is positive when the coefficient of $ \pi $ is even, while the coefficient of $ \dfrac{\pi }{6} $ is negative when the coefficient of $ \pi $ is odd.
So, after merging eq.1 and eq.2, we get
 $ \sin x = \dfrac{1}{2} $ for $ x = n\pi + {( - 1)^n}\dfrac{\pi }{6} $ where $ n \in Z $
Because, when $ n $ will have an even value, coefficient of $ \dfrac{\pi }{6} $ will become positive, and when $ n $ will have an odd value, coefficient of $ \dfrac{\pi }{6} $ will become negative.
Therefore, general solution for $ 2\sin x - 1 = 0 $ will be $ x = n\pi + {( - 1)^n}\dfrac{\pi }{6} $ where $ n \in Z $
Hence, the final solution for $ 2\cos x\sin x - \cos x = 0 $ is
 $ x = (2n + 1)\dfrac{\pi }{2} $ and $ x = n\pi + {( - 1)^n}\dfrac{\pi }{6} $ where $ n \in Z $
So, the correct answer is “ $ x = (2n + 1)\dfrac{\pi }{2} $ and $ x = n\pi + {( - 1)^n}\dfrac{\pi }{6} $ ”.

Note: To find a general solution of a trigonometric equation, first find out all the solutions in the interval after which the function repeats itself. In this case, sine and cosine functions repeat after $ 2\pi $ of interval. Thus, we found all the solutions in the range of $ (0,2\pi ) $ and then added $ 2\pi $ in the solution. Also, whenever you write a general solution, do not forget to mention where $ n $ belong. For example, in this solution, $ n \in Z $ i.e., it belongs to the set of integers.