ĐK: \(x\ne-\dfrac{\pi}{3}+k2\pi\)
\(tan\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)=2\)
\(\Leftrightarrow\dfrac{x}{2}+\dfrac{\pi}{6}=arctan2+k\pi\)
\(\Leftrightarrow x=-\dfrac{\pi}{3}+2arctan2+k2\pi\)
\(sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)tan^2x-cos^2\dfrac{x}{2}=0\)
giải phương trình
2sin3x(1-4sin2x)=1
tanx+tan(x+\(\dfrac{\pi}{3}\))+tan(x+\(\dfrac{2\pi}{3}\))=3\(\sqrt{3}\)
4(sin6x+cos6x)+2(sin4x+cos4x)=8-4cos22x
GPT sau: \(4\sin\left(x+\dfrac{\pi}{3}\right)-2\sin\left(2x-\dfrac{\pi}{6}\right)=\sqrt{3}\cos x+\cos2x-2\sin x+2\)
Giúp em giải bài này với ạ.
\(\dfrac{tan^2x+tanx}{tan^2x+1}\) = \(\dfrac{\sqrt{2}}{2}\) sin(\(\dfrac{\pi}{4}\) +x)
1) Tai sao Sin [(x+\(\dfrac{\Pi}{2}\)) - 2π] = Sin (x+\(\dfrac{\Pi}{2}\))
\(sin\dfrac{x}{2}sinx-cos\dfrac{x}{2}sin^2x+1=2cos^2\left(\dfrac{pi}{4}-\dfrac{x}{2}\right)\)
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 
giải phương trình\(\sqrt{3}cos\left(x+\dfrac{\Pi}{2}\right)+sin\left(x-\dfrac{\Pi}{2}\right)=2sin2x\)
a,cos(2x-\(\dfrac{\pi}{\text{3}}\))-4cos(x-\(\dfrac{\pi}{\text{3}}\))+3=0
b,cos x+3sin\(\dfrac{\text{x}}{\text{2}}\)-2=0
Mng giúp em với ạ, em đang cần gấp ạ. Cảm ơn mng
