\(2\sqrt{3}\sin^23x-cos6x=\sqrt{3}\\ \Leftrightarrow2sin^23x\left(\sqrt{3}+1\right)=\sqrt{3}+1\\ \Leftrightarrow sin^23x=\dfrac{1}{2}\Leftrightarrow\dfrac{1-cos6x}{2}=\dfrac{1}{2}\Rightarrow cos6x=0\Leftrightarrow x=\pm\dfrac{\pi}{12}+k\dfrac{\pi}{6}\left(k\in Z\right)\)
cosx.cos3x - sin2x.sin6x - sin4x.sin6x = 0
\(\sqrt{3}\)cos6x - 2sin4x.cos2x - sin2x = 0
giải phương trình: 2sin^2 2x-1+cos6x=0
1.2cos2x+2cosx-√2=0
2.cos6x+3sin3x+4=0
sin6x + cos6x + 3sin2xcos2x
1) Gia tri lon nhat cua ham so: y = sin6x + cos6x
Giải các pt:
a) \(cos3x-sinx=\sqrt{3}\left(cosx-sin3x\right)\)
b) \(2cos^2x-3\sqrt{3}sin2x-4sin^2x=-4\)
c) \(\sqrt{3}\left(cos2x+sin3x\right)=sin2x+cos8x\)
d) \(cos2x-\sqrt{3}sin2x=\sqrt{3}sinx+cosx\)
e) \(sin8x-cos6x=\sqrt{3}\left(sin6x+cos8x\right)\)
giải các phương trình sau: ( pt bậc nhất đối với sinx và cosx)
a, \(sinx+cosx=\sqrt{2}sin5x\)
b, \(\sqrt{3}sin2x+sin\left(\frac{\pi}{2}+2x\right)=1\)
c, \(\left(\sqrt{3}-1\right)sinx+\left(\sqrt{3}+1\right)cosx+\sqrt{3}-1=0\)
d, \(3sin^2x+\sqrt{3}sin2x=3\)
e, \(sin8x-cos6x=\sqrt{3}\left(sin6x+cos8x\right)\)
f,\(8cos2x=\frac{\sqrt{3}}{sinx}+\frac{1}{cosx}\)
g, \(cosx-\sqrt{3}sinx=2cos\left(\frac{\pi}{3}-x\right)\)
h, \(sin5x-cos5x=\sqrt{2}cos13x\)
i, \(\left(3cosx-4sinx+6\right)^2-9cosx+12sinx-16=0\)
1) Gia tri lon nhat cua ham so: y = \(\dfrac{cosx+2sinx+3}{2cosx-sinx+4}\)
A. 0 B. 3-2\(\sqrt{3}\) C. \(2-2\sqrt{2}\) D. -1
Sin4x - (sinx)^3.cosx+ 6.(sinx)^2.(cosx)^2 + 3.sinx.(cosx)^3 + 5cos4x=2
Giúp mk vs ạ cảm ơn....
