Solve the equation, \[\tan \theta +\tan 2\theta +\tan 3\theta =0\].

Hint: Use the identity, \[\tan x=\dfrac{\sin x}{\cos x}\] then take x as \[\theta ,2\theta ,3\theta \] and then so simplification and reduce it into a single trigonometric ratio.

Complete step-by-step answer:

In the question we are given an equation which is \[\tan \theta +\tan 2\theta +\tan 3\theta =0\], and we have to find the general solutions or values of \[\theta \].

So, we know that,

\[\tan \theta +\tan 2\theta +\tan 3\theta =0\]

Now, we know that, \[\tan x=\dfrac{\sin x}{\cos x}\]. So, we will apply by putting x as \[\theta ,2\theta ,3\theta \]. So, we can rewrite the equation as,

\[\dfrac{\sin \theta }{\cos \theta }+\dfrac{\sin 2\theta }{\cos 2\theta }+\dfrac{\sin 3\theta }{\cos 3\theta }=0\]

On taking L.C.M of first two terms we get,

\[\dfrac{\sin \theta \cos 2\theta +\sin 2\theta \cos \theta }{\cos \theta \cos 2\theta }+\dfrac{\sin 3\theta }{\cos 3\theta }=0\]

Now, we will apply identity, \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\] and we can use it as, \[\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)\], where, \[A=\theta \] and \[B=2\theta \]. So, we can write the equation as,

\[\dfrac{\sin 3\theta }{\cos \theta \cos 2\theta }+\dfrac{\sin 3\theta }{\cos 3\theta }=0\]

Now, on simplification we can rewrite equation as,

\[\sin 3\theta \left[ \dfrac{\cos 3\theta +\cos \theta \cos 2\theta }{\cos \theta \cos 2\theta \cos 3\theta } \right]=0\]

We can write the equation as,

\[\sin 3\theta \left( \cos 3\theta +\cos \theta \cos 2\theta \right)=0\]

Now, we will substitute and transform the terms of \[\cos 3\theta \] and \[\cos 2\theta \] in terms of \[\cos \theta \] using identities,

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

And

\[\cos 2\theta =2{{\cos }^{2}}\theta -1\]

So, we get,

\[\sin 3\theta \left( 4{{\cos }^{3}}\theta -3\cos \theta +\cos \theta \left( 2{{\cos }^{2}}\theta

-1 \right) \right)=0\]

So, on simplification we can rewrite it as,

\[\sin 3\theta \left( 4{{\cos }^{3}}\theta -3\cos \theta +2{{\cos }^{2}}\theta -\cos \theta

\right)=0\]

We can write it as,

\[\sin 3\theta \left( 6{{\cos }^{3}}\theta -4\cos \theta \right)=0\]

Or

\[2\sin 3\theta \cos \theta \left( 3{{\cos }^{3}}\theta -1 \right)=0\]

So, we can say that for \[\sin 3\theta =0,\cos \theta =0\] and \[\cos \theta =\pm \sqrt{\dfrac{2}{3}}\].

Now, let’s take the first case where, \[\sin 3\theta =0\].

We know that, \[\sin 0=0\]. So, we can write, \[\sin 3\theta =\sin 0\].

So, \[3\theta \] is equal to \[n\pi +{{\left( -1 \right)}^{n}}\left( 0 \right)\] or \[n\pi \]. Hence, \[\theta \] is equal to \[\dfrac{n\pi }{3}\], where n is an integer which we got using formula if, \[\sin x=\sin \alpha \] then, \[x=n\pi +{{\left( -1 \right)}^{n}}\alpha \], where n is an integer.

Now, let’s take the next case where, \[\cos \theta =0\].

We know that, \[\cos \dfrac{\pi }{2}=0\]. So, we can write, \[\cos \theta =\cos \dfrac{\pi }{2}\].
So, \[\theta \] is equal to \[2n\pi \pm \left( \dfrac{\pi }{2} \right)\], where n is an integer which we got by using formula, \[\cos x=\cos \alpha \]. So, \[x=2n\pi \pm \alpha \], where n is any integer.

Now, similarly for \[\cos \theta =\pm \sqrt{\dfrac{2}{3}}\].

We can write it as,

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

So, \[\theta =2n\pi \pm {{\cos }^{-1}}\left( \pm \sqrt{\dfrac{2}{3}} \right)\], where n is an integer.

Hence, the general solutions are \[n\pi \], \[2n\pi \pm \dfrac{\pi }{2}\] and \[2n\pi \pm {{\cos }^{-1}}\left( \pm \sqrt{\dfrac{2}{3}} \right)\].

Note: We can also do by using identities, \[\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }\] and \[\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3\tan \theta }\] and then changing the whole equation in terms of \[\tan \theta \].