a, 2sin[tex]^{3}[/tex]x + cos2x = sinx
b, sin3x + 2cos2x - 2 = 0
giải dùm tớ nhá
b, sin3x + 2cos2x - 2 = 0
sin3x= sin(x+2x) = 3sinx - 4sin^3x
sin3x+2cos2x-2=0
<=> 3sinx - 4sin^3x +2( 1- 2sin^2x) -2 =0
<-> 4sin^3x + 4sin^2x -3sinx = 0
<=> sinx ( 4sin^2x +4sinx -3 ) = 0