giải các phương trình
1, cos^2x+cos2x=0
2, 2cos^2x-3cos2x=4
3, sin^6x+cos^6x=cos4x/
1) ta có:
[TEX]cos^2 x+2cos^2 x-1 = 0[/TEX]
\Leftrightarrow [TEX]3cos^2 x = 1[/TEX]
\Rightarrow [TEX]cos^2 x = \frac{1}{3}[/TEX]
tự giải
2) ta có:
[TEX]2cos^2 x-3(2cos^2 x-1)=4[/TEX]
\Leftrightarrow [TEX](-4cos^2 x) = 0[/TEX]
\Rightarrow [TEX]cos^2 x = 0[/TEX]
tự giải
3) ta có:
[TEX]sin^6 x+cos^6 x = cos4x[/TEX]
\Leftrightarrow [TEX](sin^2 x+cos^2 x)^3 - 3sin^2 x.cos^2 x.(sin^2 x+cos^2 x)-cos4x = 0[/TEX]
\Leftrightarrow [TEX]1-3sin^2 x.cos^2 x-cos4x = 0[/TEX]
\Leftrightarrow [TEX]1- \frac{3}{4}sin^2 2x-cos4x = 0[/TEX]
\Leftrightarrow [TEX]1- \frac{3}{8}(1-cos4x)-cos4x = 0[/TEX]
\Leftrightarrow [TEX]\frac{3}{8}cos4x-cos4x+\frac{5}{8} = 0[/TEX]
\Leftrightarrow [TEX] (-\frac{5}{8}cos4x)+\frac{5}{8} = 0[/TEX]
\Leftrightarrow [TEX] cos4x = 1 [/TEX]
tự giải