Tìm max, min của hàm số
a) \(y=\sqrt{3}sinx+cosx\)
b) \(y=sin2x-cos2x\)
c) \(y=3sinx+4cosx\)
Tìm GTLN, GTNN:
a, \(y=\sin x+\cos x\).
b, \(y=\dfrac{1}{2}\sin x+\dfrac{\sqrt{3}}{2}\cos x+3\).
c, \(y=\sqrt{3}\sin2x-\cos2x\).
giai pt \(\dfrac{cos2x+\sqrt{3}sin2x+6sinx-5}{\dfrac{cos^2x}{2}-1}=2\sqrt{3}\)
TÌM TXĐ: y=\(\dfrac{3}{sin2x-cos2x+\sqrt{2}}\)
Tìm Min, Max của : y =\(\dfrac{4}{\sqrt{2-cos\left(x-\dfrac{\pi}{6}\right)}+3}\)
Tìm m để phương trình sau có nghiệm:
\(4sin\left(x+\dfrac{\pi}{3}\right).cos\left(x-\dfrac{\pi}{6}\right)=m^2+\sqrt{3}.sin2x-cos2x\)
Giải các pt sau
a, \(\dfrac{1}{sinx}+\dfrac{1}{cosx}=4sin\left(x+\dfrac{\pi}{4}\right)\)
b, \(2sin\left(2x-\dfrac{\pi}{6}\right)+4sinx+1=0\)
c, \(cos2x+\sqrt{3}sinx+\sqrt{3}sin2x-cosx=2\)
d, \(4sin^2\dfrac{x}{2}-\sqrt{3}cos2x=1+cos^2\left(x-\dfrac{3\pi}{4}\right)\)
Tìm GTLN, GTNN: y= sin2x + \(\sqrt{3}\)cos2x
\(\dfrac{tanx.\cos3x+2\cos2x-1}{1-2sinx}=\sqrt{3}\left(\sin2x+cosx\right)\)