3:
\(sinx-\sqrt{3}\cdot cosx=1\)
=>\(\dfrac{1}{2}\cdot sinx-\dfrac{\sqrt{3}}{2}\cdot cosx=\dfrac{1}{2}\)
=>sin(x-pi/3)=1/2
=>x-pi/3=pi/6+k2pi hoặc x-pi/3=5/6pi+k2pi
=>x=pi/2+k2pi hoặc x=4/3pi+k2pi
mà \(x\in\left[-pi;pi\right]\)
nên \(x\in\left\{\dfrac{pi}{2};-\dfrac{2}{3}pi\right\}\)
Đúng 0
Bình luận (0)
