Question

Find the general solution of $\sin 2\theta =\cos 3\theta $ .

Answer 1

Hint: Convert the whole equation in terms of sin function only, using relation $\sin \left( \dfrac{\pi }{2}-x \right)=\cos x$ with the term $\cos 3\theta $ . use the identity of $\sin C-\sin D$ to get the general solution, which is given as –

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

And use the following results as well

General solution of $\sin x=0$ is $x=n\pi $ and $\cos x=0$ is $x=\left( 2n+1 \right)\dfrac{\pi }{2}$ , $n\in z$ .

Complete step-by-step answer:
Here, we have –

$\sin 2\theta =\cos 3\theta $ ……………………………………… (i)

As the equation (i) is consisting of two different functions sin and cos both. So, we can use following trigonometric identity to get the whole equation in terms of sin only as –

$\sin \left( \dfrac{\pi }{2}-x \right)=\cos x$ …………………………. (ii)

Now, we can replace $\cos 3\theta $ by $\sin \left( \dfrac{\pi }{2}-3\theta \right)$ of the equation (i), using the equation (ii). So, we get the equation (i) as –

$\sin 2\theta =\sin \left( \dfrac{\pi }{2}-3\theta \right)$ .

Transferring the term $\sin \left( \dfrac{\pi }{2}-3\theta \right)$ to the other side of the equation, we get –

$\sin 2\theta -\sin \left( \dfrac{\pi }{2}-3\theta \right)=0$ (iii)

Now, as we know the trigonometric identity of $\sin C-\sin D$ is given as –

$\sin C-\sin D=2\sin \dfrac{C-D}{2}\cos \dfrac{C+D}{2}$ (iv)

Now, on comparing the equation (iii) and (iv), we get –

$C=2\theta $ and $D=\left( \dfrac{\pi }{2}-3\theta \right)$.

So, we can rewrite the equation (iii) with the help of equation (iv) as –

\[\begin{align}

  & 2\sin \left( \dfrac{2\theta -\left( \dfrac{\pi }{2}-3\theta \right)}{2} \right)\cos \left( \dfrac{2\theta +\left( \dfrac{\pi }{2}-3\theta \right)}{2} \right)=0 \\

 & 2\sin \left( \dfrac{2\theta -\dfrac{\pi }{2}+3\theta }{2} \right)\cos \left( \dfrac{2\theta +\dfrac{\pi }{2}-3\theta }{2} \right)=0 \\

\end{align}\]

Or \[\sin \left( \dfrac{5\theta -\dfrac{\pi }{2}}{2} \right)\cos \left( \dfrac{-\theta +\dfrac{\pi }{2}}{2} \right)=0\]

\[\sin \left( \dfrac{5\theta }{2}-\dfrac{\pi }{4} \right)\cos \left( \dfrac{-\theta }{2}+\dfrac{\pi }{4} \right)=0\]………………………………(iv)

N0w, we know that two terms in a product will be zero, if one of them will be 0. So, we can get two expressions from the equation (iv) as –

\[\sin \left( \dfrac{5\theta }{2}-\dfrac{\pi }{4} \right)=0\] or \[\cos \left( \dfrac{-\theta }{2}+\dfrac{\pi }{4} \right)=0\].

Case 1: \[\sin \left( \dfrac{5\theta }{2}-\dfrac{\pi }{4} \right)=0\]

So, the general solution of the equation \[\sin \left( \dfrac{5\theta }{2}-\dfrac{\pi }{4} \right)=0\] can be given with the help of relation (v) as –

\[\dfrac{5\theta }{2}-\dfrac{\pi }{4}=n\pi \] .

Adding \[\dfrac{\pi }{4}\] to both sides of the equation, we get –

$\begin{align}

  & \dfrac{5\theta }{2}-\dfrac{\pi }{4}+\dfrac{\pi }{4}=n\pi +\dfrac{\pi }{4} \\

 & \dfrac{5\theta }{2}=n\pi +\dfrac{\pi }{4} \\

\end{align}$

Multiplying the whole equation by ‘$\dfrac{2}{5}$’ , we get –

$\begin{align}


  & \dfrac{5\theta }{2}\times \dfrac{2}{5}=\left( n\pi +\dfrac{\pi }{4} \right)\times \dfrac{2}{5}

\\


 & \theta =\dfrac{2n\pi }{5}+\dfrac{2\pi }{20} \\


\end{align}$


Or $\theta =\dfrac{2n\pi }{5}+\dfrac{\pi }{10}$…………………………… (vi)


Where, $n\in z$ .


Case 2: \[\cos \left( \dfrac{-\theta }{2}+\dfrac{\pi }{4} \right)=0\]


Hence, the general solution of the equation \[\cos \left( \dfrac{-\theta }{2}+\dfrac{\pi }{4}


\right)=0\] by using the equation (vii). We get –


\[\dfrac{-\theta }{2}+\dfrac{\pi }{4}=\left( 2n+1 \right)\dfrac{\pi }{2}\] .


Subtracting \[\dfrac{\pi }{4}\] from the both sides of the equation, we get –


\[\begin{align}


  & -\dfrac{\theta }{2}+\dfrac{\pi }{4}-\dfrac{\pi }{4}=\left( 2n+1 \right)\dfrac{\pi


}{2}-\dfrac{\pi }{4} \\


 & -\dfrac{\theta }{2}=\left( 2n+1 \right)\dfrac{\pi }{2}-\dfrac{\pi }{4} \\


\end{align}\]


Multiplying the whole equation by ‘-2’, we can rewrite the above equation as –


\[\begin{align}

  & -\dfrac{\theta }{2}\left( -2 \right)=\left( \left( 2n+1 \right)\dfrac{\pi }{2}-\dfrac{\pi }{4}

\right)\times \left( -2 \right) \\

 & \theta =\left( 2n\times \dfrac{\pi }{2}+\dfrac{\pi }{2}-\dfrac{\pi }{4} \right)\left( -2 \right)

\\

 & \theta =\left( n\pi +\dfrac{\pi }{4} \right)\left( -2 \right) \\

\end{align}\]

\[\theta =-2n\pi -\dfrac{\pi }{2}\]…………………………………….. (viii)


Where $n\in z$

Hence, we can get the general solution of the equation $\sin 2\theta =\cos 3\theta $ , by

using the equations (vi) and (viii) as –

$\theta =\dfrac{2n\pi }{5}+\dfrac{\pi }{10},-2n\pi -\dfrac{\pi }{2}$


Where, $n\in z$.


Note: Another approach to solve the question would be given as –

$\begin{align}

  & \sin 2\theta =\cos 3\theta \\

 & \cos \left( \dfrac{\pi }{2}-2\theta \right)=\cos 3\theta \\

\end{align}$

Use a general solution for the relation $\cos x=\cos y$ i.e. $x=2n\pi \pm y$ . So, we get –

$\dfrac{\pi }{2}-2\theta =2n\pi \pm 3\theta $

Case 1:

$\begin{align}

  & \dfrac{\pi }{2}-2\theta =2n\pi \pm 3\theta \\

 & 5\theta =\dfrac{\pi }{2}-2n\pi \\

 & \theta =\dfrac{\pi }{10}-\dfrac{2n\pi }{5},n\in z. \\

\end{align}$

Case 2:

$\begin{align}

  & \dfrac{\pi }{2}-2\theta =2n\pi -3\theta \\

 & \theta =2n\pi -\dfrac{\pi }{2} \\

\end{align}$

Where, $n\in z$

So, it can be another approach. One may write the relation $\sin 2\theta =\cos 3\theta $ as $\sin 2\theta =\sin \left( \dfrac{\pi }{2}-3\theta \right)$. And hence, use the general solution for $\sin x=\sin y$ , which is given as –

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

Where, \[n\in z\].

So, one may observe that the representation of the solution of the same equation is different but the solutions on putting values of ‘n’ will be the same. Hence, do not confuse, if your answer is not matching with the given solution, just put the values of n and hence, try to verify your relation.