Question

Find the general solution of $\sin 2\theta =\dfrac{\sqrt{3}}{2}$ .

Answer 1

Hint: Get the angle at which the sine function will give a value of $\dfrac{\sqrt{3}}{2}$ . 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 $\sin {{60}^{\circ }}$ or $\sin \left( \dfrac{\pi }{3} \right)$ is $\dfrac{\sqrt{3}}{2}$ ;

use this result to get the general solution of the given equation.

Complete step-by-step answer:

Given expression in the problem is –

$\sin 2\theta =\dfrac{\sqrt{3}}{2}$………………… (i)

As we know the trigonometric function $\sin \theta $ will give value $\dfrac{\sqrt{3}}{2}$ at $\theta ={{60}^{\circ }}$ or we can convert to ${{30}^{\circ }}$ to radian and hence, we get $\sin \theta $ will give $\dfrac{\sqrt{3}}{2}$ at $\dfrac{\pi }{6}\left( \pi =180{}^\circ \right)$ .
So, we can replace $\dfrac{\sqrt{3}}{2}$ in the right hand side of the expression in the equation (i), by $\sin \dfrac{\pi }{3}$ .

So, we can rewrite the equation (i) as –

  $\sin 2\theta =\dfrac{\sqrt{3}}{2}$ ……………………….. (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 $\sin 2\theta =\dfrac{\sqrt{3}}{2}$ with equation $\sin x=\sin y$, we get –
$x=2\theta $ and $y=\dfrac{\pi }{3}$.

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

$2\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}$ .

On dividing the whole equation by 2, we get –

\[\dfrac{2\theta }{2}=\dfrac{n\pi }{2}+{{\left( -1 \right)}^{n}}\dfrac{\pi }{12}\]

Or \[\theta =\dfrac{n\pi }{2}+{{\left( -1 \right)}^{n}}\dfrac{\pi }{12}\].

Where, we can put values of n (integer), to get the corresponding number of solutions. So, the general solution of the equation $\sin 2\theta =\dfrac{\sqrt{3}}{2}$ is given as –

\[\theta =\dfrac{n\pi }{2}+{{\left( -1 \right)}^{n}}\dfrac{\pi }{12}\].



Note: One may go wrong if he/she uses trigonometric relation of $\sin 2\theta $ , given as –
$\sin 2\theta =2\sin \theta \cos \theta $ .

Or, students may use another relation to replace $\sin 2\theta $ as well, but while applying these relations, the given trigonometric relation will become more complex. So, the given relation is in simplest form, do not use any other identity to make the relation complex.

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

$\sin x-\sin y=0$ .

Apply $\sin C-\sin D=2\sin \dfrac{C-D}{2}\cos \dfrac{C+D}{2}$ .

So, we get –

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

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