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[snowangel1103]

$1/  2+\sqrt{3}(sin2x-3sinx)=cos2x+3cosx$

 

2/ $\frac{4\sqrt{3}sinx.cos^2x-2cos\frac{5x}{2}cos\frac{x}{2}+\sqrt{3}sin2x+3cosx+2}{2sinx-\sqrt{3}}=0$

Edited by snowangel1103, 09-05-2013 - 22:16.

[tranphuonganh97]

câu 1:

phương trình đề cho tương đương:

$(cos2x-\sqrt{3}sin2x)+3(cosx+\sqrt{3}sinx)=2 <=>(\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x)+3(\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx)=1 <=> cos(2x+\frac{\Pi }{3})+3sin(x+\frac{\Pi }{6})-1=0 <=> 1-2sin^2(x+\frac{\Pi }{6})+3sin(x+\frac{\Pi }{6})-1=0 <=> -2sin^2(x+\frac{\Pi }{6})+3sin(x+\frac{\Pi }{6})=0.$

đến đây dễ giải tiếp.

Edited by tranphuonganh97, 11-05-2013 - 18:09.