Phương trình lượng giác 11

N

nhoka3

đề như vầy hả bạn
$\frac{sin3x}{sinx}$$-tan^2x-$$\frac{2}{cos2x}$$=1$

\Leftrightarrow $\frac{3sinx-4sin^3x}{sinx}$$-(tan^2x+1)-$$\frac{2}{cos2x}$$=0$

\Leftrightarrow $3-4sin^2x-$$\frac{1}{cos^2x}$$-$$\frac{2}{2cos^2x-1}$$=0$

\Leftrightarrow $4cos^2x-1-$$\frac{2cos^2x-1+2cos^2x}{cos^2x(2cos^2x-1)}$$=0$

\Leftrightarrow $(4cos^2x-1)cos^2x(2cos^2x-1)-4cos^2x+1=0$

\Leftrightarrow $4cos^6x-5cos^4x-3cos^2x+1=0 (1)$

đặt $t= cos^2x$ >0$

(1)\Leftrightarrow $4t^3-5t^2-3t+1=0$

\Leftrightarrow $(4t-1)(t^2-t-1)=0$
bạn giải típ nha
 
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