[phương trình lượng giác]

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xuanquynh97

ĐK $tanx \not=\dfrac{-3}{2}; cosx \not=0$

PT \Leftrightarrow $5+ [cos^2x - sin^2x = 6.cosx + 4.cosx.\frac{sinx}{cosx}$

\Leftrightarrow $cos^2x- 6.cosx + 9 = sin^2x+ 4.sinx + 4$

\Leftrightarrow $(cosx-3)^2= (sinx+2)^2$

\Leftrightarrow $3-cosx = sinx + 2$

\Leftrightarrow $cosx + sinx = 1$
 
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