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L

locxoaymgk

[TEX]PT \Leftrightarrow \frac{3}{2}sinx+\frac{1}{2}cosx=\frac{-1}{4}[/TEX]

[TEX] \Leftrightarrow sin(x+\frac{\pi}{6})=-1/4[/TEX]

[TEX]\Leftrightarrow \left{\begin{x+\frac{\pi}{6}=arcsin-1/4+k2\pi}\\{x+\frac{\pi}{6}=\pi-arcsin-1/4+k2\pi}[/TEX]
 
N

nguyenbahiep1

L

locxoaymgk

locxoaymgk said:
[TEX]PT \Leftrightarrow \frac{3}{2}sinx+\frac{1}{2}cosx=\frac{-1}{4}[/TEX]

[TEX] \Leftrightarrow sin(x+\frac{\pi}{6})=-1/4[/TEX]

[TEX]\Leftrightarrow \left{\begin{x+\frac{\pi}{6}=arcsin-1/4+k2\pi}\\{x+\frac{\pi}{6}=\pi-arcsin-1/4+k2\pi}[/TEX]
Bấm để xem đầy đủ nội dung ...

Đặt [TEX]cosa=-1/4.[/TEX]

[TEX]PT \Leftrightarrow sin(x+\frac{\pi}{6})=cosa[/TEX]

[TEX]\Leftrightarrow cos(\frac{\pi}{3}-x)=cosa[/TEX]

[TEX]\Rightarrow ...[/TEX]
 
Y

youaremysoul

pt \Leftrightarrow $4(\dfrac{\sqrt{3}}{2}sinx + \dfrac{1}{2}cosx )= -1$

\Leftrightarrow $4cos(\dfrac{pi}{3} - x) = -1$

\Leftrightarrow $cos(\dfrac{pi}{3} - x) = \dfrac{-1}{4}$

\Rightarrow ..................................
 
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Y

youaremysoul

tam_1996 said:
\Rightarrow.......................................................... (Cái gì)
trả lời đầy đủ giúp @-) nha bạn
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thì làm như pt bt
đặt $cosa = \dfrac{-1}{4}$

\Rightarrow $cos(\dfrac{pi}{3} - x) =cosa$

\Leftrightarrow [TEX]\left[\begin{$\dfrac{pi}{3} - x = a +k2pi$}\\{$\dfrac{pi}{3} - x = -a +k2pi$} [/TEX]

\Leftrightarrow [TEX]\left[\begin{$x = \dfrac{pi}{3} - a - k2pi$}\\{$x = \dfrac{pi}{3} + a - k2pi$} [/TEX]
 
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