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.