Đề như này pk?
\(\dfrac{1+sinx}{1+cosx}=\dfrac{1}{2}\) (đk:\(cosx\ne-1\Leftrightarrow x\ne\pi+k2\pi\left(k\in Z\right)\))
\(\Leftrightarrow2\left(1+sinx\right)=1+cosx\)
\(\Leftrightarrow2sinx-cosx=-1\)
\(\Leftrightarrow\dfrac{2}{\sqrt{5}}sinx-\dfrac{1}{\sqrt{5}}cosx=-\dfrac{1}{\sqrt{5}}\)
Đặt \(cos\alpha=\dfrac{2}{\sqrt{5}}\Rightarrow sin\alpha=\dfrac{1}{\sqrt{5}}\) (vì \(cos^2\alpha+sin^2\alpha=1\))
\(\Rightarrow sin\left(-\alpha\right)=-\dfrac{1}{\sqrt{5}}\)
\(\Rightarrow sinx.cos\alpha-cosx.sin\alpha=sin\left(-\alpha\right)\)
\(\Leftrightarrow sin\left(x-\alpha\right)=sin\left(-\alpha\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\alpha=-\alpha+k2\pi\\x-\alpha=\pi+\alpha+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi+2\alpha+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)(thỏa mãn)
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