phương trình bậc nhất đối với sin, cos : sin(2x+9pi/2)-3cos(x-3pi/2)=1+2sinx 03/10/2021 Bởi Mary phương trình bậc nhất đối với sin, cos : sin(2x+9pi/2)-3cos(x-3pi/2)=1+2sinx
$$\eqalign{ & \sin \left( {2x + {{9\pi } \over 2}} \right) – 3\cos \left( {x – {{3\pi } \over 2}} \right) = 1 + 2\sin x \cr & \Leftrightarrow \sin \left( {2x + {\pi \over 2} + 4\pi } \right) – 3\cos \left( {x + {\pi \over 2} – 2\pi } \right) = 1 + 2\sin x \cr & \Leftrightarrow \sin \left( {2x + {\pi \over 2}} \right) – 3\cos \left( {x + {\pi \over 2}} \right) = 1 + 2\sin x \cr & \Leftrightarrow \cos 2x + 3\sin x = 1 + 2\sin x \cr & \Leftrightarrow \sin x = 1 – \cos 2x \cr & \Leftrightarrow \sin x = 2{\sin ^2}x \cr & \Leftrightarrow \sin x\left( {2\sin x – 1} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ \sin x = 0 \hfill \cr \sin x = {1 \over 2} \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = k\pi \hfill \cr x = {\pi \over 6} + k2\pi \hfill \cr x = {{5\pi } \over 6} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$ Bình luận
$$\eqalign{
& \sin \left( {2x + {{9\pi } \over 2}} \right) – 3\cos \left( {x – {{3\pi } \over 2}} \right) = 1 + 2\sin x \cr
& \Leftrightarrow \sin \left( {2x + {\pi \over 2} + 4\pi } \right) – 3\cos \left( {x + {\pi \over 2} – 2\pi } \right) = 1 + 2\sin x \cr
& \Leftrightarrow \sin \left( {2x + {\pi \over 2}} \right) – 3\cos \left( {x + {\pi \over 2}} \right) = 1 + 2\sin x \cr
& \Leftrightarrow \cos 2x + 3\sin x = 1 + 2\sin x \cr
& \Leftrightarrow \sin x = 1 – \cos 2x \cr
& \Leftrightarrow \sin x = 2{\sin ^2}x \cr
& \Leftrightarrow \sin x\left( {2\sin x – 1} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
\sin x = 0 \hfill \cr
\sin x = {1 \over 2} \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = k\pi \hfill \cr
x = {\pi \over 6} + k2\pi \hfill \cr
x = {{5\pi } \over 6} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$