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conga222222

heomoon said:
[TEX]\frac{sinx + cosx}{sinx - cosx} + 2tan2x + cos2x = 0[/TEX]
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\[\begin{array}{l}
\frac{{sinx + cosx}}{{sinx - cosx}} + 2tan2x + cos2x = 0(1)\\
dk...\\
(1) \leftrightarrow \frac{{\left( {\sin x + \cos x} \right)\cos 2x}}{{\sin x - \cos x}} + 2\sin 2x + {\cos ^2}2x = 0\\
\leftrightarrow \frac{{\left( {\sin x + \cos x} \right)\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}}{{\sin x - \cos x}} + 2\sin 2x + {\cos ^2}2x = 0\\
\leftrightarrow - {\left( {\sin x + \cos x} \right)^2} + 2\sin 2x + {\cos ^2}2x = 0\\
\leftrightarrow - \left( {1 + 2\sin 2x} \right) + 2\sin 2x + 1 - {\sin ^2}2x = 0\\
\leftrightarrow - {\sin ^2}2x = 0
\end{array}\]
 
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