Question

Solve the equation \[3{{\cos }^{2}}\theta -2\sqrt{3}\sin \theta \cos \theta -3{{\sin }^{2}}\theta =0\].

Answer 1

Hint: We will apply the formula \[\cos 2\theta =co{{s}^{2}}\theta -{{\sin }^{2}}\theta \] and also the trigonometric formula given by \[\sin 2\theta =2\sin \theta \cos \theta \]. We will also use the inverse trigonometric formula \[{{\tan }^{-1}}\left( \tan \theta \right)=\theta \] to simplify and solve the given equation.

Complete step-by-step answer:

Considering the equation \[3{{\cos }^{2}}\theta -2\sqrt{3}\sin \theta \cos \theta -3{{\sin }^{2}}\theta =0...(1)\]

We will first rewrite this equation (1) by replacing \[-2\sqrt{3}\sin \theta \cos \theta \] and \[-3{{\sin }^{2}}\theta \] with each other. Thus, we get the expression as,

\[3{{\cos }^{2}}\theta -3{{\sin }^{2}}\theta -2\sqrt{3}\sin \theta \cos \theta =0\]

Now we will take the ‘3’ number as common from the term \[3{{\cos }^{2}}\theta -3{{\sin }^{2}}\theta \]. Therefore we have,

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

We will also write \[\sqrt{3}.2\] instead of \[2\sqrt{3}\] in the above expression. Thus we get,
\[3\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)-\sqrt{3}.2\sin \theta \cos \theta =0\].

In this step we will apply two trigonometric formulas which are given by \[\cos 2=co{{s}^{2}}\theta -{{\sin }^{2}}\theta \] and \[\sin 2\theta =2\sin \theta \cos \theta \].
Therefore, we get,

\[3\cos 2\theta -\sqrt{3}.sin2\theta +\sqrt{3}\sin 2\theta =\sqrt{3}\sin 2\theta \].

In this step we can clearly see that the terms, \[-\sqrt{3}sin2\theta \] and \[\sqrt{3}\sin 2\theta \] gets cancelled. Therefore we have a new expression, given as,

\[3\cos 2\theta =\sqrt{3}\sin 2\theta \].

Divide the expression by \[\sqrt{3}\] on both the sides. Thus we get,

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

Now we will cancel \[\sqrt{3}\] from numerator and denominator both and we will do this on both the sides of the above equation. Thus, we have that \[\sqrt{3}\cos 2\theta =\sin 2\theta \]. Now we will divide the expression by \[\cos 2\theta \]. Therefore, we get,

\[\dfrac{\sqrt{3}\cos 2\theta }{\cos 2\theta }=\dfrac{\sin 2\theta }{\cos 2\theta }\] or \[\sqrt{3}=\dfrac{\sin 2\theta }{\cos 2\theta }......(2)\]

In trigonometry, the value of \[\dfrac{\sin x}{\cos x}=\tan x\].

By considering \[x=\theta \], we get \[\sqrt{3}=\tan 2\theta \] by equation (2).

Now we will apply the inverse operation here on both the sides. So we get a new expression as \[{{\tan }^{-1}}\left( \sqrt{3} \right)={{\tan }^{-1}}\left[ \tan \left( 2\theta \right) \right]\].
Here we choose the inverse tangent trigonometric expression,

\[{{\tan }^{-1}}\left( \sqrt{3} \right)={{\tan }^{-1}}\left[ \tan \left( 2\theta \right) \right]\].
By the formula \[{{\tan }^{-1}}\left[ \tan \left( 2\theta \right) \right]=2\theta \] we get a new expression, given by \[{{\tan }^{-1}}\left( \sqrt{3} \right)=2\theta \]. Now we will apply the formula or we can say just the value of \[\left( \sqrt{3} \right)\] in trigonometric terms. Here, \[\left( \sqrt{3} \right)\] can be written as \[\tan \dfrac{\pi }{3}\]. So by this substitution, we have,

\[{{\tan }^{-1}}\left( \tan \dfrac{\pi }{3} \right)=2\theta \].

By the formula, \[{{\tan }^{-1}}\left( \tan \theta \right)=\theta \], we get \[{{\tan }^{-1}}\left( \tan \dfrac{\pi }{3} \right)=\dfrac{\pi }{3}\].

\[\begin{align}

  & \dfrac{\pi }{3}=2\theta \\

 & \theta =\dfrac{\pi }{6} \\

\end{align}\]

Hence the given solution of equation (1) is \[\theta =\dfrac{\pi }{6}\].


Note: We could have applied \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] or \[\cos 2\theta =1-2{{\sin }^{2}}\theta \] but there is no need in this equation \[3{{\cos }^{2}}\theta -2\sqrt{3}\cos \theta \sin \theta -3\sin 2\theta \] since we directly got the answer that \[\theta =\dfrac{\pi }{6}\] so there is no question about the general form of the equation’s solution.