\(sin^4\left(x+\dfrac{\pi}{2}\right)-sin^4x=sin4x\)
\(\Rightarrow cos^4x-sin^4x=sin4x\)
\(\Rightarrow\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=sin4x\)
\(\Rightarrow cos^2x-sin^2x=4sinx.cosx.cos2x\)
......
\(sin^4\left(x+\dfrac{\pi}{2}\right)-sin^4x=sin4x\)
\(\Rightarrow cos^4x-sin^4x=sin4x\)
\(\Rightarrow\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=sin4x\)
\(\Rightarrow cos^2x-sin^2x=4sinx.cosx.cos2x\)
......
Giải các phương trình lượng giác:
a) \(sin4x-cos\left(x+\dfrac{\pi}{6}\right)=0\)
b) \(cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)
c) \(cos4x=cos\dfrac{5\pi}{12}\)
d) \(cos^2x=1\)
Pt \(tan\left(\dfrac{\pi}{4}.sin4x\right)=\dfrac{3}{2}\) có bao nhiêu họ nghiệm?
giải các phương trình sau : a) \(\sin4x=\sin\frac{\pi}{5}\) ; b) \(\sin\left(\frac{x+\pi}{5}\right)=-\frac{1}{2}\) ; c) \(\cos\frac{x}{2}=\cos\sqrt{2}\) ; d) \(\cos\left(x+\frac{\pi}{18}\right)=\frac{2}{5}\)
1/ tìm x thuộc (\(\dfrac{-\pi}{2}\);\(\dfrac{\pi}{2}\)) sao cho tan(3x+2)=\(\sqrt{3}\)
2/ tìm x thuộc (0; 3\(\pi\)) sao cho sin(x-\(\dfrac{\pi}{3}\))+2cos(x+\(\dfrac{\pi}{6}\))=0
Giúp em với mọi người :)
1) \(\frac{1}{\cos x}+\frac{1}{\sin2x}=\frac{2}{\sin4x}\)
2) \(\cos3x\cdot\tan5x=\sin7x\)
3) \(\tan5x\cdot\tan2x=1\)
4) \(4\cos x-2\cos2x-\cos4x=1\)
5) \(\sin\left(2x+\frac{5\pi}{2}\right)-2\cos\left(x-\frac{7\pi}{2}\right)=1+2\sin x\)
6) \(\sin^22x-\cos^28x=\sin\left(\frac{17\pi}{2}+10x\right)\)
7) \(8\cos x=\frac{\sqrt{3}}{\sin x}+\frac{1}{\cos x}\)
Giải các pt sau:
a) \(\sin\left(3x+60^o\right)=\dfrac{1}{2}\)
b) \(\cos\left(2x-\dfrac{\pi}{3}\right)=\dfrac{-\sqrt{2}}{2}\)
c) \(\tan\left(x+\dfrac{\pi}{6}\right)=\sqrt{3}\)
d) \(\cot\left(2x+\pi\right)=-1\)
giải các pt sau
1) \(\sqrt{2}\sin^3(x+\dfrac{\pi}{4})=2\sin x\)
2) \(\sin^3\left(x-\dfrac{\pi}{4}\right)=\sqrt{2}\sin x\)
Giải phương trình:
\(\sin\left(\dfrac{3\pi}{10}-\dfrac{x}{2}\right)=\dfrac{1}{2}\sin\left(\dfrac{\pi}{10}+\dfrac{3x}{2}\right)\)
Giải các phương trình :
a) \(\tan\left(2x+45^0\right)=-1\)
b) \(\cot\left(x+\dfrac{\pi}{3}\right)=\sqrt{3}\)
c) \(\tan\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)=\tan\dfrac{\pi}{8}\)
d) \(\cot\left(\dfrac{x}{3}+20^0\right)=-\dfrac{\sqrt{3}}{3}\)