Giải phương trình sau : a) cos2x.tanx=0 b) sin3x.cotx=0 04/10/2021 Bởi Madeline 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) Bình luận
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) ™\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} $ Bình luận
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) ™\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} $