Question

 

Answer 1

a: \(\Leftrightarrow\sqrt{6}\cdot sin\left(2x-\dfrac{pi}{3}\right)=\sqrt{3}\)

=>\(sin\left(2x-\dfrac{pi}{3}\right)=\dfrac{1}{\sqrt{2}}=sin\left(\dfrac{pi}{4}\right)\)

=>2x-pi/3=pi/4+k2pi hoặc 2x-pi/3=-pi/4+k2pi

=>2x=7/12pi+k2pi hoặc 2x=-1/12pi+k2pi

=>x=7/24pi+kpi hoặc x=-1/24pi+kpi

b: \(\Leftrightarrow sin2x\cdot\dfrac{\sqrt{3}}{2}-cos2x\cdot\dfrac{1}{2}=sin3x\)

=>sin(2x-pi/3)=sin3x

=>3x=2x-pi/3+k2pi hoặc 3x=pi-2x+pi/3+k2pi

=>3x-2x=-pi/3+k2pi hoặc 3x+2x=4/3pi+k2pi

=>x=-pi/3+k2pi hoặc x=4/15pi+k2pi/5