Question

How do you solve $ \sin x\cos x = \dfrac{1}{2} $ ?

Answer 1

Hint: In order to solve the above trigonometric equation, multiply both sides with the number 2 and use write $ 2\sin x\cos x $ as $ \sin 2x $ using the trigonometric identity . Take the inverse of sine on both sides and find the generalised solution of $ x $ by using the fact that the period of sine function is $ 2\pi $ .

Complete step by step solution:
We are given a trigonometric equation $ \sin x\cos x = \dfrac{1}{2} $ .
 $ \sin x\cos x = \dfrac{1}{2} $
Multiplying both sides of the equation with the number $ 2 $ ,
 $
  2\sin x\cos x = \dfrac{1}{2} \times 2 \\
  2\sin x\cos x = 1 \;
  $
Using the identity of trigonometry of sine double angle i.e. $ \sin 2x = 2\sin x\cos x $ .So replacing $ 2\sin x\cos x $ with $ \sin 2x $ in the above equation , we have
 $ \sin 2x = 1 $
Now taking inverse of sine function on both sides of the equation
 $ {\sin ^{ - 1}}\left( {\sin 2x} \right) = {\sin ^{ - 1}}\left( 1 \right) $
Since $ {\sin ^{ - 1}}\left( {\sin } \right) = 1 $ as they both are inverse of each other
 $ 2x = {\sin ^{ - 1}}\left( 1 \right) $
 $ {\sin ^{ - 1}}\left( 1 \right) $ is nothing but an angle whose sine is equal to 1. As we know $ \sin \dfrac{\pi }{2} = 1 \to \dfrac{\pi }{2} = {\sin ^{ - 1}}\left( 1 \right) $ . But if we try to generalise the solution, we have a period of sine function $ 2\pi $ as the sine function repeats itself after every $ 2\pi $ interval.
Hence,
 $ 2x = 2n\pi + \dfrac{\pi }{2} $
Dividing both side by 2 , we get
\[
  \dfrac{{2x}}{2} = \dfrac{1}{2}\left( {2n\pi + \dfrac{\pi }{2}} \right) \\
  x = n\pi + \dfrac{\pi }{4} \;
 \]
Where n is an integer
Therefore, the solution to the given trigonometric equation is \[x = n\pi + \dfrac{\pi }{4}\]
So, the correct answer is “\[x = n\pi + \dfrac{\pi }{4}\]”.

Note: 1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2.The answer obtained should be in a generalised form.
3. The domain of sine function is in the interval $ \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ and the range is in the interval $ \left[ { - 1,1} \right] $ .