Phương trình lượng giác.

Question

Giải phương trình: $\left(\tan x+1\right)\sin^2x+\cos2x+2= 3\left(\cos x+\sin x\right)\sin x$

score 3 · Answer 1

$(tanx+1)sin^2x+cos2x+2=3(cosx+sinx)sinx$

$\Leftrightarrow cos2x+2-2(cosx+sinx)sinx=(cosx+sinx)sinx-(tanx+1)sin^2x$

$\Leftrightarrow cos2x+2-2cosxsinx-2sin^2x=(cosx+sinx)sinx(1-\frac{sinx}{cosx})$

$\Leftrightarrow cos2x+2cos^2x-2cosxsinx=(cosx+sinx)sinx.\frac{cosx-sinx}{cosx}$

$\Leftrightarrow (cos^2x-sin^2x)+2cosx(cosx-sinx)=(cosx+sinx).\frac{sinx}{cosx}.(cosx-sinx)$

$\Leftrightarrow (cosx-sinx)(3cosx+sinx)=(cosx-sinx).(cosx+sinx).\frac{sinx}{cosx}$

TH1.

$cosx-sinx=0$

$\Rightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+2k\pi , \frac{5\pi}{4}+2k\pi$

TH2.

$3cosx+sinx=(cosx+sinx).\frac{sinx}{cosx}$

$\Leftrightarrow (3cosx+sinx)cosx=(cosx+sinx)sinx$

$\Leftrightarrow 3cos^2x=sin^2x$

$\Leftrightarrow tan^2x=3$

$\Leftrightarrow tanx=\pm \sqrt3$

$\Leftrightarrow x=\frac{\pi}{3}+2k\pi , \frac{2\pi}{3}+2k\pi , \frac{4\pi}{3}+2k\pi , \frac{5\pi}{3}+2k\pi$