Question

Solve the following equation:
${\tan ^2}2\theta + 2\tan 3\theta \tan 2\theta - 1 = 0$

Answer 1

Hint : In this particular question use the concept of some basic trigonometric identities such as, tan x = (sin x/cos x), $\cos 2x = {\cos ^2}x - {\sin ^2}x$, $2\sin x\cos x = \sin 2x$, $\cot x = \tan \left( {\dfrac{{ \pm \pi }}{2} - x} \right)$ so use these identities to reach the solution of the question.

Complete step by step solution :
Given trigonometric equation is
${\tan ^2}2\theta + 2\tan 3\theta \tan 2\theta - 1 = 0$
Now this equation is also written as
$\Rightarrow 2\tan 3\theta \tan 2\theta = 1 - {\tan ^2}2\theta$
Now as we know that tan x = (sin x/cos x), so use this property in the above equation we have,
$\Rightarrow 2\dfrac{{\sin 3\theta }}{{\cos 3\theta }}\dfrac{{\sin 2\theta }}{{\cos 2\theta }} = 1 - \dfrac{{{{\sin }^2}2\theta }}{{{{\cos }^2}2\theta }}$
$\Rightarrow 2\dfrac{{\sin 3\theta }}{{\cos 3\theta }}\dfrac{{\sin 2\theta }}{{\cos 2\theta }} = \dfrac{{{{\cos }^2}2\theta - {{\sin }^2}2\theta }}{{{{\cos }^2}2\theta }}$
Now cancel out the common terms from the denominator of both sides of the above equation, and use the property that, $\cos 2x = {\cos ^2}x - {\sin ^2}x \Rightarrow \cos 4x = {\cos ^2}2x - {\sin ^2}2x$ so we have,
$\Rightarrow 2\dfrac{{\sin 3\theta }}{{\cos 3\theta }}\sin 2\theta = \dfrac{{\cos 4\theta }}{{\cos 2\theta }}$
Now take $2\sin 2\theta$ from the L.H.S to the denominator of the R.H.S we have,
$\Rightarrow \dfrac{{\sin 3\theta }}{{\cos 3\theta }} = \dfrac{{\cos 4\theta }}{{2\sin 2\theta \cos 2\theta }}$
Now as we know that, $2\sin x\cos x = \sin 2x \Rightarrow 2\sin 2x\cos 2x = \cos 4x$, so use this property in the above equation we have,
$\Rightarrow \dfrac{{\sin 3\theta }}{{\cos 3\theta }} = \dfrac{{\cos 4\theta }}{{\sin 4\theta }}$
Now as we know that, (sin x/cos x) = tan x and (cos x/sin x) = cot x, so use this property in the above equation we have,
$\Rightarrow \tan 3\theta = \cot 4\theta$
Now as we know that, $\cot x = \tan \left( {\dfrac{{ \pm \pi }}{2} - x} \right)$, so use this property in the above equation we have,
$\Rightarrow \tan 3\theta = \tan \left( {\dfrac{{ \pm \pi }}{2} - 4\theta } \right)$
Now cancel out tan from both sides we have,
$\Rightarrow 3\theta = \left( {\dfrac{{ \pm \pi }}{2} - 4\theta } \right)$
$\Rightarrow 7\theta = \dfrac{{ \pm \pi }}{2}$
$\Rightarrow \theta = \dfrac{{ \pm \pi }}{{14}}$
$\Rightarrow \theta = \dfrac{\pi }{{14}}, - \dfrac{\pi }{{14}}$
So this is the required solution of the given trigonometric equation.
So this is the required answer.

Note : Whenever we face such types of questions the key concept we have to remember is that always recall the basic trigonometric identities which is the basis of the above solution and which is all stated above, so first simplify the equation using these identities as above we will get the required solution of the given equation.