Question

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

Answer 1

Hint: In order to determine the value of the above question, use the trigonometric table to find the angle in the interval $ \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] $ for which sine is 0.5 to get the required result.

Complete step-by-step answer:
Given $ \sin x = \dfrac{1}{2} $
 $ x = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) $
We know that $ {\sin ^{ - 1}}\theta $ denotes an angle in the interval $ \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] $ whose sine is $ x $ for $ x \in \left[ { - 1,1} \right]. $
Therefore,
 $ x = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) $ = An angle in $ \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] $ , whose sine is $ \dfrac{1}{2} $ .
From the trigonometric table we have,
 $ \sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2} $
Transposing sin from left-hand side to right-hand side
 $ x = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6} $
Hence, the required answer is $ x = \dfrac{\pi }{6} $
So, the correct answer is “ $ x = \dfrac{\pi }{6} $ ”.

Note: 2. In inverse trigonometric function, the domain are the ranges of corresponding trigonometric functions and the range are the domain of the corresponding trigonometric function.
3. 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.
If T is the smallest positive real number such that $ f(x + T) = f(x) $ for all x, then T is called the fundamental period of $ f(x) $ .
Since $ \sin \,(2n\pi + \theta ) = \sin \theta $ for all values of $ \theta $ and n $ \in $ N.
4. 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.