Giải phương trình sau : a) cos2x.tanx=0 b) sin3x.cotx=0

a) cos2x.tanx=0

ĐK: cosx khác 0 ⇒x khác pi/2 +kpi

cos2x .sinx=0

⇒cos2x=0 ⇒x=pi/4 +kpi/2 (TM)

hoặc sinx=0 ⇒x=kpi (TM)

b) sin3x.cotx=0

ĐK : sinx khác 0 ⇒x khác kpi

sin3x .cosx =0

⇒sin3x=0 ⇒x=kpi/3 (TM)

hoặc cosx=0 ⇒x=pi/2+kpi (TM)

Đáp án:

1) $\left\{ \matrix{ x = {\pi \over 4} + {{k\pi } \over 2} \hfill \cr x = k\pi \hfill \cr} \right. (k\in\mathbb Z)$

2) $\left\{ \matrix{ x = {\pi \over 3} + k\pi \hfill \cr x = {\pi \over 2} + k\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right)$

Lời giải:

$\eqalign{ & a)\,\,\cos 2x\tan x = 0 \cr & \text{Đkxđ: }x \ne {\pi \over 2} + k\pi \cr & pt \Leftrightarrow \left[ \matrix{ \cos 2x = 0 \hfill \cr \sin x = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ 2x = {\pi \over 2} + k\pi \hfill \cr x = k\pi \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = {\pi \over 4} + {{k\pi } \over 2} \hfill \cr x = k\pi \hfill \cr} \right.\,\,\left( {k\in\mathbb Z} \right) (tm)\cr & b)\,\,\sin 3x\cot x = 0 \cr & \text{Đkxđ: }\sin x \ne 0 \Leftrightarrow x \ne k\pi \cr & pt \Leftrightarrow \left( {3\sin x - 4{{\sin }^3}x} \right){{\cos x} \over {\sin x}} = 0 \cr & \Leftrightarrow \left( {3 - 4{{\sin }^2}x} \right)\cos x = 0 \cr & \Leftrightarrow \left( {3 - 2\left( {1 - \cos 2x} \right)} \right)\cos x = 0 \cr & \Leftrightarrow \left( {2\cos 2x + 1} \right)\cos x = 0 \cr & \Leftrightarrow \left[ \matrix{ \cos 2x = - {1 \over 2} \hfill \cr \cos x = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ 2x = {{2\pi } \over 3} + k2\pi \hfill \cr x = {\pi \over 2} + k\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = {\pi \over 3} + k\pi \hfill \cr x = {\pi \over 2} + k\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right)\,\,\left( {tm} \right) \cr}$