Giai phương trình:
1. cos^2x+cos^2(2x)+cos^2(3x)-1=0
2. sin^3x+cos^3x-2(sin^5x+cos^5x)=0
Help me,please!!
Lần sau bạn lập topic mới nhé
1.
[TEX]cos^{2}x+cos^{2}2x+cos^{2}3x-1=0[/TEX]
=> [TEX]\frac{1+cos2x}{2}+\frac{1+cos4x}{2}+cos^{2}3x-1=0[/TEX]
=> [TEX]cos2x+cos4x+2cos^{2}3x=0[/TEX]
=> [TEX]2cosx.cos3x+2cos^{2}3x=0[/TEX]
=> [TEX]cos3x(cosx+cos3x)=0[/TEX]
=> ......................
2.
[TEX]sin^{3}x+cos^{3}x-2(sin^{5}x+cos^{5}x)=0[/TEX]
=> [TEX]sin^{3}x(1-2sin^{2})-cos^{3}(2cos^{2}x-1)=0[/TEX]
=> [TEX]sin^{3}x.cos2x-cos^{3}x.cos2x=0[/TEX]
=> [TEX](sin^{3}-cos^{3}).cos2x=0[/TEX]
=> [TEX]cos2x(sinx-cosx)(sin^{2}x+sinx.cosx+cos^{2}x)=0[/TEX]
=> [TEX]cos2x(sinx-cosx)(1+\frac{sin2x}{2})=0[/TEX]
=> [TEX]cos2x=0[/TEX] hoặc [TEX]sinx=cosx[/TEX] hoặc [TEX]1+\frac{sin2x}{2}=0[/TEX]
+) cos2x=0
=> [tex]2x=\pm \frac{k\pi }{2}+k2\pi \Rightarrow x=\pm \frac{k\pi }{4}+k\pi[/tex]
+) sinx=cosx
=> [TEX]sinx=sin(\frac{\pi }{2}-x)[/TEX]
=> [TEX]x=\frac{\pi }{2}-x+k2\pi [/TEX] hoặc [TEX]x=\frac{\pi }{2}+x+k2\pi [/TEX]
=> ..............
+) [TEX]1+\frac{sin2x}{2}=0[/TEX]
=> sin2x=-2 (loại)