Question

Find the general solution of $\sin \theta =\dfrac{1}{2}$.

Answer 1

Hint: Get the angle at which the sine function will give a value of $\dfrac{1}{2}$. Write the general solution of equation $\sin x=\sin y$, by following result:

\[x=n\pi +{{\left( -1 \right)}^{n}}y\]

Where \[n\in z\].

Value of $\sin {{30}^{\circ }},\sin \left( \dfrac{\pi }{6} \right)\to \dfrac{1}{2}$; use this result to get the general solution of the given equation.

Complete step-by-step answer:


Given expression in the problem is

$\sin \theta =\dfrac{1}{2}$ …………….(i)

As we know the trigonometric function $\sin \theta $ will give value $\dfrac{1}{2}\to \theta ={{30}^{\circ }}$or we can convert ${{30}^{\circ }}$ to radian and hence, we get $\sin \theta \to \dfrac{1}{2}\to \dfrac{\pi }{6}\left( \pi ={{180}^{\circ }} \right)$.

So, we can replace $\dfrac{1}{2}$ in the right hand side of the expression in the equation (i), by $\sin \left( \dfrac{\pi }{6} \right)$ . so, we can rewrite the equation (i) as

$\sin \theta =\sin \left( \dfrac{\pi }{6} \right)$ ……………..(ii)

Now, as we know the general solution of equation $\sin x=\sin y$ can be given as
If $\sin x=\sin y$, then

$x=n\pi +{{\left( -1 \right)}^{n}}y$ …………………(iii)

Where, \[n\in z\]i.e. n is an integer.

Now, we can compare the equation (ii) with the equation $\sin x=\sin y$and hence, we can get values of x and y, so that the general solution can be calculated with the equation (iii). So, on comparing $\sin \theta =\sin \left( \dfrac{\pi }{6} \right)$ with the equation $\sin x=\sin y$, we get

$x-\theta ,y=\dfrac{\pi }{6}$

Now, we can substitute these values to the equation (iii) to get the general solution of equation (i). so, we get

$\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}$

Where, \[n\in z\].

Now, we can put a number of values of n to get solutions.

Note: One may prove the general solution of $\sin x=\sin y$by following ways:

$\sin x-\sin y=0$

Apply $\operatorname{sinC}-\operatorname{sinD}=2sin\dfrac{C-D}{2}\cos \dfrac{C+D}{2}$
So, we get

$2\sin \dfrac{x-y}{2}\cos \dfrac{x+y}{2}=0$

Now, put $\sin \left( \dfrac{x-y}{2} \right)=0,\cos \left( \dfrac{x+y}{2} \right)=0$

And apply the general solution of equations $\sin \theta =0,\cos \theta =0$ which are $\theta =n\pi ,\theta =\left( 2n+1 \right)\dfrac{\pi }{2}$ respectively. Hence, get the general solution for $\sin x=\sin y$

One may take any other value of $\dfrac{1}{2}$ in equation $\sin \theta =\dfrac{1}{2}$ other

than $\dfrac{\pi }{6}$ as there are infinite angles at which $\sin $ will give $\dfrac{1}{2}$.

But by the conventional approach, we need to take the angle from $0\to \pi $ only.

So, putting any other ‘y’ in the general solution (other than $\dfrac{\pi }{6}$ ) will not affect the solutions of the given equation. But, by convection, we put ‘y’ from 0 to $\pi $ only. So, take care of it.