[Toán 11] Phương trình lượng giác

0

0ocxinh

Hình như đề sai hay sao ấy
[tex]%20cos^{2}x%20+%20cos^{2}2x%20+%20cos^{2}3x%20+%20cos^{2}4x%20=\frac{3}{2}[/tex]

thế này mới ra dc
 
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C

connguoivietnam

[TEX]cos^2x+cos^22x+cos^23x+cos^24x=\frac{3}{2}[/TEX]

[TEX]2cos^2x+2cos^22x+2cos^23x+2cos^24x=3[/TEX]

[TEX]1+cos2x+1+cos4x+1+cos6x+1+cos8x=3[/TEX]

[TEX]cos2x+cos4x+cos6x+1+cos8x=0[/TEX]

[TEX]2cos4xcos2x+cos4x+2cos^24x=0[/TEX]

[TEX]cos4x(2cos2x+1+2cos4x)=0[/TEX]

[TEX]cos4x(2cos2x+4cos^22x-1)=0[/TEX]
 
0

0ocxinh

[tex]%20cos^{2}x%20+%20cos^{2}2x%20+%20cos^{2}3x%20+%20%20%20cos^{2}4x%20=\frac{3}{2}[/tex]
\Leftrightarrow [tex]\frac{1+cos2x}{2}+\frac{1+cos4x}{2}+\frac{1+cos6x}{2}+%20%20%20cos^{2}4x%20=\frac{3}{2}[/tex]
\Leftrightarrow [tex]\frac{(cos2x+cos6x)+cos4x}{2}+%20%20%20cos^{2}4x=0[/tex]
\Leftrightarrow [tex]\frac{2cos4xcos2x+cos4x}{2}+%20%20%20cos^{2}4x%20=0[/tex]
\Leftrightarrow [tex](2cos4xcos2x+cos4x+2%20%20%20cos^{2}4x%20=0[/tex]
\Leftrightarrow [TEX]cos4x(2cos4x+2cos2x+1)=0[/TEX]
\Leftrightarrow [TEX]cos4x(4%20cos^{2}2x%20+2cos2x+1)=0[/TEX]
\Leftrightarrow cos4x=0
hoặc cos2x= (-1 + [TEX]\sqrt{5}[/TEX] trên 4
hoặc cos2x=(1 + [TEX]\sqrt{5}[/TEX] trên 4
\Leftrightarrow [tex]\left[\begin{x= \frac{/pi}{8}+ \frac{/kpi}{4}}\\{x = \pm \frac{/pi}{5} + kpi }\\{x=\pm \frac{2pi}{5}+kpi} [/tex]
 
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