First slide

Trigonometric equations

Question

If $\tan 3 θ + \tan θ = 2 \tan 2 θ,$ then $θ$ is equal to $(n \in Z)$

Moderate

A

$nπ$
B

$\frac{nπ}{4}$
C

$\frac{nπ}{2}$
D

none of these

Solution

$\tan 3 θ + \tan θ = 2 \tan 2 θ,$
$\begin{array}{l} or \tan 3 θ - \tan 2 θ = \tan 2 θ - \tan θ \\ or \frac{\sin (3 θ - 2 θ)}{\cos 3 θcos 2 θ} = \frac{\sin (2 θ - θ)}{\cos 2 θcos θ} \\ or \sin θ (2 \sin θsin 2 θ) = 0 \\ or \sin θ = 0 or \sin 2 θ = 0 \\ \Rightarrow θ = nπ or 2 θ = nπ, n \in Z \end{array}$
But $θ = nπ / 2$ is rejected because when n is odd, tan $θ$ is not defined and when n is even, i.e., 2r, then $θ = rπ$
Then $\dot{θ} = nπ, n \in I$ is the only solution

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