Question

bài 1 

\(\text{a: (sinx-cosx)}^2=2-2cos^2x\)

b:\(tanx+\sqrt{3}cotx-\sqrt{3}-1=0\)

c:\(sinx-sin2x+sin3x=0\)

d: \(\dfrac{3}{cos^2x}+\left(\sqrt{3}-3\right)tanx=3+\sqrt{3}\)

Answer 1

a: \(\Leftrightarrow1-sin2x-2+2\cdot\dfrac{1+cos2x}{2}=0\)

=>-1-sin2x+1+cos2x=0

=>sin2x-cos2x=0

=>căn 2 sin(2x-pi/4)=0

=>2x-pi/4=kpi

=>2x=kpi+pi/4

=>x=kpi/2+pi/8

b:

ĐKXĐ: x<>pi/2+kpi và x<>kpi

 \(\Leftrightarrow tanx+\dfrac{\sqrt{3}}{tanx}-\sqrt{3}-1=0\)

=>\(tan^2x+\sqrt{3}+tanx\left(-\sqrt{3}-1\right)=0\)

\(\Leftrightarrow tanx\left(tanx-\sqrt{3}\right)-tanx+\sqrt{3}=0\)

=>tanx=căn 3 hoặc tanx=1

=>x=pi/4+kpi hoặc x=pi/3+kpi

c: \(\Leftrightarrow2\cdot sinx\cdot sin2x-sin2x=0\)

=>sin2x(2sinx-1)=0

=>sin2x=0 hoặc sinx=1/2

=>\(\left[{}\begin{matrix}x=\dfrac{kpi}{2}\\x=\dfrac{pi}{6}+k2pi\\x=\dfrac{5}{6}pi+k2pi\end{matrix}\right.\)