Question

How do you solve $ \sin \left( {\pi + x} \right) = 0.5 $ between $ 0 < x < 2\pi $ ?

Answer 1

Hint: In order to determine the solution of the above trigonometric equation in the interval $ 0 < x < 2\pi $ . Since we know that the sine function is always positive in the 1st and 2nd quadrant. So there will be two solutions to the given equation. For the first quadrant use the property $ {\sin ^{ - 1}}\left( {\sin x} \right) = x $ and for the 2nd quadrant use $ {\sin ^{ - 1}}\left( {\sin x} \right) = \pi - x $ property of inverse sine.

Complete step by step solution:
We are given a trigonometric equation $ \sin \left( {\pi + x} \right) = 0.5 $ and we have to find the solution between $ 0 < x < 2\pi $
 $ \sin \left( {\pi + x} \right) = 0.5 $
Rewriting the above equation , by taking inverse of sine on both sides , we get
 $ {\sin ^{ - 1}}\left( {\sin \left( {\pi + x} \right)} \right) = {\sin ^{ - 1}}\left( {0.5} \right) $ ----(1)
Since $ {\sin ^{ - 1}}\left( {\sin } \right) = 1 $ as they both are inverse of each other
 $ \pi + x = {\sin ^{ - 1}}\left( {0.5} \right) $
As we know the sine of angle $ \dfrac{\pi }{6} $ is equal to 0.5 in the first quadrant.
 $
  \pi + x = \dfrac{\pi }{6} \\
  x = \dfrac{\pi }{6} - \pi \;
  $
Combining like terms , we have
 $
  x = \dfrac{{\pi - 6\pi }}{6} \\
  x = \dfrac{{ - 5\pi }}{6} \;
  $
Since the sine function is positive in 1st and 2nd quadrant both, so using the property of inverse sine function that $ {\sin ^{ - 1}}\left( {\sin x} \right) = \pi - x $ where $ x \in \left[ {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right] $ in the equation (1), we get
 $
  {\sin ^{ - 1}}\left( {\sin \left( {\pi + x} \right)} \right) = {\sin ^{ - 1}}\left( {0.5} \right) \\
  \pi + x = \pi - \dfrac{\pi }{6} \\
  x = - \dfrac{\pi }{6} \;
  $
Therefore, the solution to the given trigonometric equation is \[x = - \dfrac{\pi }{6}, - \dfrac{{5\pi }}{6}\] between $ 0 < x < 2\pi $
So, the correct answer is “\[x = - \dfrac{\pi }{6}, - \dfrac{{5\pi }}{6}\]”.

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] $ .