[lớp 11] Giải phương trình lượng giác $2sin2x - cos2x=7sinx+2cosx -4$

H

heodat_15

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N

nguyenbahiep1

a, 2sin2x - cos2x=7sinx+2cosx -4
b,sinx+2cosx +cos2x -2sinxcosx=0
c,tan(2x -15^o)=1 với -pi < x<pi/2

câu a

[TEX]4sinx.cosx - 2cosx - 1 + 2sin^2x - 7sinx + 4 = 0 \\ 2cosx(2sinx-1) + 2sin^2x -7sinx +3 \\ 2cosx(2sinx-1) + (2sinx-1)(sinx-3) = 0 \\ \Rightarrow sin x = \frac{1}{2} \\ 2 cosx + sin x = 3 (v/n)[/TEX]

câu c

[TEX]tan (2x - \frac{\pi}{12}) = tan (\frac{\pi}{4}) \\ 2x - \frac{\pi}{12} = \frac{\pi}{4} + k.\pi \\ k : Z \\ x = \frac{\pi}{6} + \frac{k.\pi}{2} \\ - \frac{\pi}{2} \leq \frac{\pi}{6} + \frac{k.\pi}{2} \leq \frac{\pi}{2} \\ - \frac{1}{2} \leq \frac{1}{6}+ \frac{k}{2} \leq \frac{1}{2} \\ - \frac{4}{3} \leq k \leq \frac{2}{3} \Rightarrow k = 0 , k = -1 \\ x = \frac{\pi}{6} \\ x = - \frac{\pi}{3}[/TEX]
 
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Y

youaremysoul

b,sinx+2cosx +cos2x -2sinxcosx=0

\Leftrightarrow $sinx + 2cosx + 1 - 2sin^2x - 2sinxcosx = 0$

\Leftrightarrow $(-2sin^2x + sinx + 1) + 2cosx(1 - sinx) = 0$

\Leftrightarrow $(1 - sinx)(2sinx + 1) + 2cosx(1 - sinx) = 0$

\Leftrightarrow $(1 - sinx)[2(sinx + cosx) + 1] = 0 $
 
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