Question

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