Tìm nghiệm của các phương trinh:
1,\(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
2,\(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}\left(1+cot2xcotx\right)=0\)
3,\(cos^4x+sin^4x+cos\left(x-\dfrac{\pi}{4}\right)sin\left(3x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)
4,\(cos5x+cos2x+2sin3xsin2x=0\) trên \(\left[0;2\pi\right]\)
5,\(\dfrac{cos\left(cosx+2sinx\right)+3sinx\left(sinx+\sqrt{2}\right)}{sin2x-1}=1\)
6,\(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
7,\(cos\left(2x+\dfrac{\pi}{4}\right)+cos\left(2x-\dfrac{\pi}{4}\right)+4sinx=2+\sqrt{2}\left(1-sinx\right)\)
Nghiệm của phương trình \(sin^4x+cos^4x+cos\left(x-\dfrac{\pi}{4}\right).sin\left(3x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)
giải phương trình
\(\sin x\sqrt{1+2\sin x}=\cos2x\)
\(\sin\left(\frac{5x}{2}-\frac{\pi}{4}\right)-\cos\left(\frac{x}{2}-\frac{\pi}{4}\right)=\sqrt{2}\cos\frac{3x}{2}\)
\(3\sqrt{\tan x+1}\left(\sin x+2\cos x\right)=5\left(\sin x+3\cos x\right)\)
\(\sqrt{2}\left(\sin x+\sqrt{3}\cos x\right)=\sqrt{3}\cos2x-\sin2x\)
\(\sin2x\sin4x+2\left(3\sin x-4\sin^2x+1\right)=0\)
giải các pt
a) \(cos^2x+sin2x-1=0\)
b) \(\sqrt{3}sin2x+\:cos^4x-sin^4x=\sqrt{2}\)
c) \(\:cos^2x-sin^2x=\sqrt{2}.sin\left(x+\frac{\pi}{4}\right)\)
d) \(4\left(sin^4x+cos^4x\right)+\sqrt{3}.sin4x=2\)
e) \(4sinx.cosx.cos2x+cos4x=\sqrt{2}\)
sin3x=\(\frac{-\sqrt{3}}{2}\)
\(sin\left(2x-\frac{\pi}{7}\right)=\frac{\sqrt{2}}{2}\)
\(sin\left(4x+1\right)=\frac{3}{5}\)
\(sin\left(2x+\frac{\pi}{7}\right)=sin\left(x-\frac{3\pi}{7}\right)\)
\(sin\left(4x+\frac{\pi}{7}\right)=\frac{1}{4}\)
MỌI NGƯỜI GIÚP MÌNH VỚI MÌNH CẢM ƠN NHIỀU
\(\dfrac{2sin^3x+2\sqrt{3}sin^2x.cosx-2sin^2x+cos\left(2x+\dfrac{\pi}{3}\right)}{2cosx-\sqrt{3}}=0\)
\(\sin\left(3x+\frac{\pi}{4}\right)+8\sin^2x-\sqrt{2}\sin x=2\)
giải phương trình
a, \(2\sin\frac{x}{2}\left(\sin\frac{3x}{2}+\cos\frac{3x}{2}\right)=3-4\cos x\)
b, \(\frac{2\cos^2x+\sqrt{3}\sin2x+3}{2\cos^2x.\sin\left(x+\frac{\pi}{3}\right)}=3\left(\tan^2x+1\right)\)
\(2\sin^2\left(5\pi+1\right)-\left(\sqrt{3}+1\right)\sin2\left(\frac{\pi}{2}-x\right)+\sqrt{3}\sin^2\left(\frac{9\pi}{2}+x\right)=0\)
giải các pt
a) \(sin^3x.cosx-sinx.cos^3x=\frac{\sqrt{2}}{8}\)
b) \(sin^3x-cos^24x=sin^25x-cos^26x\)
c) \(\left(2sinx-cosx+1\right)\left(1+cosx\right)=sin^2x\)
d) \(sin7x+sin9x=2\left[cos^2\left(\frac{\pi}{4}-x\right)-cos^2\left(\frac{\pi}{4}+2x\right)\right]\)