Question

Solve the equation, \[\sin 3\theta -\sin \theta =4{{\cos }^{2}}\theta -2\].

Answer 1

Hint: At first use identity \[2{{\cos }^{2}}\theta -1=\cos 2\theta \] and change right hand side of equation and then use identity, \[\sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)\] to transform left hand side and hence proceed to find what is asked.

Complete step-by-step answer:

In the question we are given that an equation is \[\sin 3\theta -\sin \theta =4{{\cos }^{2}}\theta -2\] and we have to find general solutions or values of \[\theta \].

So, we are given that,

\[\sin 3\theta -\sin \theta =4{{\cos }^{2}}\theta -2\]

We can rewrite it as,

\[\sin 3\theta -\sin \theta =2\left( {{\cos }^{2}}\theta -1 \right)\]

Now, as we know that, \[2{{\cos }^{2}}\theta -1=\cos 2\theta \]. So, we can transform equation as,

\[\sin 3\theta -\sin \theta =2\cos 2\theta \].

Now, we will apply formula which is,

\[\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)\]

So, instead of A we will put \[3\theta \] and instead of B we will put \[\theta \] so we get,

\[2\cos \left( \dfrac{3\theta +\theta }{2} \right)\sin \left( \dfrac{3\theta -\theta }{2} \right)=2\cos \theta \]

Or

\[2\cos \theta \sin \theta =2{{\cos }^{2}}\theta \]

So, we can take \[2{{\cos }^{2}}\theta \] from right to left hand side so we get,

\[2\cos \theta \sin \theta -2\cos 2\theta =0\]

Or

\[2\cos \theta \left( \sin \theta -1 \right)=0\]

So, to satisfy the equation we can say that either \[2\cos \theta =0\] or \[\sin \theta =0\].

Let’s take \[{{1}^{st}}\] case where, \[\cos 2\theta =0\]. We know that, \[\cos \dfrac{\pi }{2}=0\]. So, we write, \[2\theta =2n\pi \pm \dfrac{\pi }{2}\] or \[\theta =n\pi \pm \dfrac{\pi }{4}\], where n is any integer using formula, if \[\cos \theta =\cos \alpha \], where n is any integer.

Now let’s take another case where, \[\sin \theta =1\].

We know that, \[\sin \dfrac{\pi }{2}=1\]. So, we can write that,

\[\sin \theta =\sin \dfrac{\pi }{2}\]

So, we can write, \[\theta =n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{2} \right)\], where n is any integer using formula, if \[\sin \theta =\sin \alpha \], then \[\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha \], where n is any integer.

Hence, general solutions are \[n\pi \pm \dfrac{\pi }{4}\] and \[n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{2} \right)\].

Note: Students can also solve by changing \[\sin 3\theta \] as in form of \[\sin \theta \] by using identity, \[\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta \] and \[{{\cos }^{2}}\theta \] as \[1-{{\sin }^{2}}\theta \]. Then take \[\sin \theta \] as x and find values of x. Hence find \[\theta \] which will become a very lengthy and hectic process.