[toán 11] c

N

nguyenbahiep1

câu 4

[TEX] -3cosx + 4cos^3x -4(2cos^2x-1) + 3cosx - 4 = 0 \\ 4cos^3x -8cos^2x = 0 \\ cosx = 0 \\ cosx = 2(L) \\ cosx = 0 \Rightarrow x = \frac{\pi}{2} + k.\pi \\ 0 \leq \frac{\pi}{2} + k.\pi \leq 14 \Rightarrow -\frac{1}{2} \leq k \leq \frac{14}{\pi} - \frac{1}{2} \\ \Rightarrow k = 0,1,2,3 \Rightarrow x = \frac{\pi}{2} \\ x = \frac{3.\pi}{2} \\ x= \frac{5.\pi}{2} \\ x = \frac{7.\pi}{2}[/TEX]
 
N

newstarinsky

$1) 3cosx+cos2x-cos3x+1=2sinx.sin2x\\
\Leftrightarrow 3cosx+cos2x-cos3x+1=cosx-cos3x\\
\Leftrightarrow 2cosx+2cos^2x=0\\
\Leftrightarrow cosx(cosx+1)=0$

$2) cosx+cos3x-cos2x=0\\
\Leftrightarrow 2cos2x.cosx-cos2x=0\\
\Leftrightarrow cos2x(2cosx-1)=0$

$3) sin^23x-cos^24x=sin^25x-cos^26x\\
\Leftrightarrow (sin3x-sin5x)(sin3x+sin5x)=(cos4x-cos6x)(cos4x+cos6x)\\
\Leftrightarrow -cos4x.sinx.sin4x.cosx=sinx.sin5x.cosx.cos5x\\
\Leftrightarrow sinx.cosx(sin10x+sin8x)=0$
 
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