0
0samabinladen


Giải phương trình:
1.ĐH Công Đoàn
[TEX]\frac{sin^2x-2}{sin^2x-4cos^2(\frac{x}{2})} = tan^2(\frac{x}{2})[/TEX]
2.ĐH KTQD:
[TEX]cosxcos2xcos4xcos8x = \frac{1}{16}[/TEX]
3.ĐHDHN:
[TEX]\frac{cos(\frac{4x}{3})-cos^2x}{\sqrt{1-tan^2x}} = 0[/TEX]
4.ĐHBKHN:
[TEX]\frac{1}{tanx+cotan2x} = \frac{\sqrt{2}(cosx-sinx)}{cotanx-1}[/TEX]
5.HVNH:
[TEX]2+cosx = 2tan^2(\frac{x}{2})[/TEX]
6.ĐHQGHN:
[TEX]2tanx+cotan2x = 2sin2x +\frac{1}{sin2x}[/TEX]
Last edited by a moderator: