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...
Đọc tiếp
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 
\(cos3x=sin\left(x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow cos3x=cos\left(\dfrac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{4}-x+k2\pi\\3x=-\dfrac{\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)
Pt\(\Rightarrow cos3x=cos[\dfrac{\pi}{2}-(x+\dfrac{\pi}{4})]\)
\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{2}-(x+\dfrac{\pi}{4})+k2\pi\\3x=-\dfrac{\pi}{2}+\left(x+\dfrac{\pi}{4}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{4}+k2\pi\\2x=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+k2\pi\\x=-\dfrac{\pi}{8}+k2\pi\end{matrix}\right.\)(k\(\in\)Z)