\(VT=sin^4x\cdot\dfrac{cos^2x}{sin^2x}+cos^4x\cdot\dfrac{sin^2x}{cos^2x}+sin^4x-sin^2x\cdot cos^2x\)
\(=sin^2x\cdot cos^2x+cos^2x\cdot sin^2x+sin^4x-sin^2x\cdot cos^2x\)
\(=sin^2x\left(sin^2x+cos^2x\right)=sin^2x=VP\)
\(VT=sin^4x\cdot\dfrac{cos^2x}{sin^2x}+cos^4x\cdot\dfrac{sin^2x}{cos^2x}+sin^4x-sin^2x\cdot cos^2x\)
\(=sin^2x\cdot cos^2x+cos^2x\cdot sin^2x+sin^4x-sin^2x\cdot cos^2x\)
\(=sin^2x\left(sin^2x+cos^2x\right)=sin^2x=VP\)
Giúp mình với !!!
Giải pt sau :
√3.cotx = 8cos2x.cosx - 1
sin2x.cos2x = sin7x.cos4x
2cosx.(sin2x+sinx - 1) = 1
giải pt 2sinx + cos3x +sin2x=1+sin4x
Giải phương trình :
\(2\sin6x-2\sin4x+\sqrt{3}\cos2x=\sqrt{3}+\sin2x\)
GPT
a) \(\left(2sinx-1\right)\left(\sqrt{3}cosx-5\right)=0\)
b) \(sin2x.cos2x.cos4x+\frac{1}{8}=0\)
c) \(sin4x+\sqrt{3}sin2x=0\)
d) \(\left(\sqrt{2}sin2x+2\right)\left(2cosx+\sqrt{2}\right)=0\)
tìm min max của hàm số sau
\(y=\left(2cos^2x-\sqrt{3}sin4x\right)-3\left(sin2x-\sqrt{3}cos2x\right)+3\)
Thu gọn biểu thức:
A=sin2x+sin4x+sin6x+sin8x
B=\(\frac{sin2a-2sin4a+sin6a}{1+cos2a+cos4a}\)
C=\(\frac{cos5a.cos3a+sin7a.sina}{sin6a+sin2a}\)
Tìm gtnn, gtln của hs y=sin2x.cos2x
Tìm GTLN và GTNN:
1.\(y=\sqrt{5-2cos^2x.sin^2x}\)
2.\(y=1+\dfrac{1}{2}sin2x.cos2x\)
3.\(y=\sqrt{1+sinx}-3\)
4.\(y=\sqrt{2+sin^22x}\)
sin4x+sqrt(3)cos4x = 2