Question

Find the general solution of the equation $ \sin 2x + \cos x = 0 $

Answer 1

Hint: In order to solve this question we will first break the $ \sin 2x $ function and then we will find the common from these two as it is given in question we will put it equal to zero then we will find two factors and then from these we will find the value of x in radians.

Complete step by step solution:
For solving this question we will split the $ \sin 2x $ function though identity of $ \sin 2\theta = 2\sin \theta \cos \theta $
So as in question it was given that:
 $ \sin 2x + \cos x = 0 $
The new expression will be:
 $ 2\sin x\cos x + \cos x = 0 $
From expression we will take the common $ \cos x $ so the new expression will be transformed as:
 $ \cos x(2\sin x + 1) = 0 $
As we can see there are two factors so we will put:
 $ \cos x = 0 $ or $ 2\sin x + 1 = 0 $
From the first factor $ \cos x = 0 $ the value of x will be equal to $ \dfrac{\pi }{2} $
So the general solution will be:
 $ x = \left( {2n + 1} \right)\dfrac{\pi }{2},n \in Z $
And from the second factor the value of $ \sin x = - \dfrac{1}{2} $
Now on finding the value where we are getting the value on $ \sin x = - \dfrac{1}{2} $ is $ \dfrac{{5\pi }}{6} $
So the general solution will be:
 $ x = n\pi + {\left( { - 1} \right)^n}\dfrac{{7\pi }}{6},n \in Z $
So the general solution of this expression will be
 $ x = \left( {2n + 1} \right)\dfrac{\pi }{2} $ or $ x = n\pi + {\left( { - 1} \right)^n}\dfrac{{7\pi }}{6},n \in Z $
So, the correct answer is “ $ x = \left( {2n + 1} \right)\dfrac{\pi }{2} $ or $ x = n\pi + {\left( { - 1} \right)^n}\dfrac{{7\pi }}{6},n \in Z $ ”.

Note: While solving these types of problems we should always keep in mind that we should not substitute one value to the other side if we are doing it many times there is possibility that we miss one value of the solutions and sometimes we will not be able to find even one solution also.