giải phương trình\(\sqrt{3}cos\left(x+\dfrac{\Pi}{2}\right)+sin\left(x-\dfrac{\Pi}{2}\right)=2sin2x\)
GPT sau: \(4\sin\left(x+\dfrac{\pi}{3}\right)-2\sin\left(2x-\dfrac{\pi}{6}\right)=\sqrt{3}\cos x+\cos2x-2\sin x+2\)
a, cos4x + 12sin2x -1 = 0
b, cos4x - sin4x + cos4x = 0
c, 5.(sinx + \(\dfrac{cos3x+sin3x}{1+2sin2x}\) ) = 3 + cos2x với mọi x\(\in\left(0;2\pi\right)\)
d, \(\dfrac{sin3x}{3}=\dfrac{sin5x}{5}\)
e, \(\dfrac{sin5x}{5sinx}=1\)
f, cos23x - cos2x - cos2x =0
g, cos4x + sin4x + cos(\(x-\dfrac{\pi}{4}\) ) . sin(\(3x-\dfrac{\pi}{4}\) ) - \(\dfrac{3}{2}\) = 0
h, sin\(\left(2x+\dfrac{5\pi}{2}\right)\) - 3cos\(\left(x-\dfrac{7\pi}{2}\right)\)= 1 + 2sinx với x\(\in\left(\dfrac{\pi}{2};2\pi\right)\)
i, 5sinx - 2 = 3.( 1- sinx ) . tan3x
k, ( sin2x + \(\sqrt{3}cos2x\))2 - 5 = cos \(\left(2x-\dfrac{\pi}{6}\right)\)
l, \(\dfrac{2.\left(cos^6x+sin^6x\right)-sinx.cosx}{\sqrt{2}-2sinx}=0\)
m, \(\dfrac{\left(1+sinx+cos2x\right).sin\left(x+\dfrac{\pi}{4}\right)}{1+tanx}=\dfrac{1}{\sqrt{2}}cosx\)
Mọi người giúp mình nha ! Mình cần gấp cho ngày mai
\(cosx-2cos3x=1+\sqrt{3}sinx\)
\(sinx+sinx\left(x+\dfrac{\pi}{3}\right)+sin4x=sin\left(2x-\dfrac{\pi}{3}\right)\)
\(\left(1-\dfrac{1}{2sinx}\right)cos^22x=2sinx-3+\dfrac{1}{sinx}\)
( sinx -2cosx)cos2x + sinx = (cos4x - 1)cosx +\(\dfrac{cos2x}{2sinx}\)
\(\left(\dfrac{cos4x+sin2x}{cos3x+sin3x}\right)^2=2\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+3\)
GPT: \(\dfrac{4\sin^2\dfrac{x}{2}-\sqrt{3}\cos2x-1-2\cos^2\left(x-\dfrac{3\pi}{4}\right)}{\sqrt{2\cos3x+1}}=0\)
Giải phương trình:
\(sin3x-cos3x-\sqrt{2}sin\left(5x-\dfrac{\pi}{3}\right)=0\)
cos2x-√3 sin2x=sin3x+1
3sin2x+4cos2x+5cos2003x=0
√3sin(x-\(\frac{\pi}{3}\))\(+sin\left(x+\frac{\pi}{6}\right)-2sin1972x=0\)
\(\sqrt{2}cos\left(\frac{x}{5}-\frac{\pi}{12}\right)-\sqrt{6}sin\left(\frac{x}{5}-\frac{\pi}{12}\right)=2sin\left(\frac{x}{5}+\frac{2\pi}{3}\right)-2sin\left(\frac{3x}{5}+\frac{\pi}{6}\right)\)
\(\dfrac{\sqrt{2}\left(sinx-cox\right)^2\left(1+2sin2x\right)}{sin3x+sin5x}=1-tanx\)
\(sin\left(2x-\dfrac{\pi}{4}\right)cos2x-2\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=0\)
(sin2x+cos2x)cosx+2cos2x -sinx=0
sinx + cosxsin2x + \(\sqrt{3}cos3x=2\left(cos4x+sin^3x\right)\)
\(\sqrt{3}cos5x-2sin3xcos2x-sinx=0\)
giải phương trình sau
\(\sqrt{3}sinx+cosx+2cos\left(x-\dfrac{\pi}{3}\right)=2\)