Question

Solve $3tan2x-4tan3x=ta{n}^{2}3xtan2x$

• A. $x=n\pi ,$ $n\pi \ne ta{n}^1\sqrt{3/5}\forall nϵz$
• B. $x=n\pi ,$ $n\pi \ne co{t}^1\sqrt{3/5}\forall nϵz$
• C. $x=n\pi /2,$ $n\pi \ne ta{n}^1\sqrt{3/5}\forall nϵz$
• D. $x=n\pi ,$ $n\pi \ne ta{n}^1\sqrt{2/5}\forall nϵz$

Answer 1

The correct option is A $x=n\pi ,$ $n\pi \ne ta{n}^1\sqrt{3/5}\forall nϵz$

$3\tan 2x-4\tan 3x={\tan }^{2}3x.\tan 2x$

$\Rightarrow 4\tan 2x-4\tan 3x=\tan 2x+{\tan }^{2}2x.{\tan }^{2}x$

for $\cos 2x.\cos 3x\ne 0$

$\frac{-4\sin x}{\cos 2x.\cos 3x}=\tan 2x(1+{\tan }^{2}x)$

$\Rightarrow \frac{-4\sin x}{\cos 2x.\cos 3x}=\frac{2\sin x.\cos x}{\cos 2x.{\cos }^{2}3x}$

$\Rightarrow \sin x=0or\frac{\cos x}{\cos 3x}=-2$

$\therefore x=n\pi$, $n\pi \ne {\tan }^{-1}\sqrt{\frac{3}{5}}$