1 + sinx + sinx.cosx + 2cosx - cosx.sin^2x = 0
<=.> 1 + sinx + 2sinx.cosx + 2cosx - cosx.sin^2x - sinxcosx = 0
<=> 1 + sinx + 2cosx(sinx+1) - sinxcosx(sinx+1) = 0
<=> (1+sinx)( 1+ 2cosx- sinxcosx)=0
<=> 1+sinx =0 (1)
hoặc 1+ 2cosx- sinxcosx =0 (2)
(1) <=> sinx =-1 => x = 3π/2 +k2π
(2) <=> 1+ 2cosx- sinxcosx =0
<=> sin^2(x/2) +cos^2(x/2)+2[cos^2(x/2)−sin^2(x/2))-[2sin(x/2)cos(x/2)(cos^2(x/2)−sin^2(x/2)]=0
đặt a=x/2
<=> 3cos^2a - sin^2a - 2cos^3a.sina + 2sin^3a.cosa=0
<=> 3cos^2a (sin^2a+ cos^2a) - sin^2a (sin^2a+ cos^2a) - 2cos^3a.sina + 2sin^3a.cosa=0
<=> sin^4a - 2cos^2a. sin^2 a - 3cos^4a + 2cos^3a.sina - 2sin^3a.cosa =0
xét cosa = 0 <=> cos(x/2) = 0<=> x/2 = π/2 =kπ => x= π/4 + kπ/2
xét cosa khác 0, chia 2 vế pt cho cos^4a
<=> tan^4a - -2tan^3a-2tan^2a+2tanx -3 =0
giải pt bậc 4 của tan(a) rồi đưa ra nghiệm.