Giải phương trình \(\frac{\sin ^{10} x+\cos ^{10} x}{4}=\frac{\sin ^{6} x+\cos ^{6} x}{4 \cos ^{2} 2 x+\sin ^{2} 2 x}\)
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Lời giải:
Báo saiTa có: \(4 \cos ^{2} 2 x+\sin ^{2} 2 x=3 \cos ^{2} 2 x+1>0, \forall x \in \mathbb{R}\)
\(\begin{array}{l} \frac{\sin ^{10} x+\cos ^{10} x}{4}=\frac{\sin ^{6} x+\cos ^{6} x}{4 \cos ^{2} 2 x+\sin ^{2} 2 x} \Leftrightarrow \frac{\sin ^{10} x+\cos ^{10} x}{4}=\frac{\sin ^{6} x+\cos ^{6} x}{4\left(\cos ^{2} x-\sin ^{2} x\right)^{2}+4 \sin ^{2} x \cdot \cos ^{2} x} \\ \Leftrightarrow \frac{\sin ^{10} x+\cos ^{10} x}{4}=\frac{\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cdot \cos ^{2} x+\cos ^{4} x\right)}{4\left(\cos ^{4} x-\sin ^{2} x \cdot \cos ^{2} x+\cos ^{4} x\right)} \\ \Leftrightarrow \sin ^{10} x+\cos ^{10} x=1(1) \end{array}\)
Ta có \(\left\{\begin{array}{l} \sin ^{10} x \leq \sin ^{2} x \\ \cos ^{10} x \leq \cos ^{2} x \end{array} \Rightarrow \sin ^{10} x+\cos ^{10} x \leq \sin ^{2} x+\cos ^{2} x=1\right.\)
\((1) \Leftrightarrow \left\{ {\begin{array}{*{20}{l}} {{{\sin }^{10}}x = {{\sin }^2}x}\\ {{{\cos }^{10}}x = {{\cos }^2}x} \end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}} \begin{array}{l} \left[ \begin{array}{l} {\sin ^2}x = 1\\ {\sin ^2}x = 0 \end{array} \right.\\ \left[ \begin{array}{l} {\cos ^2}x = 1\\ {\cos ^2}x = 0 \end{array} \right. \end{array} \end{array} \Leftrightarrow \left[ {\begin{array}{*{20}{l}} {{{\sin }^2}x = 0}\\ {{{\cos }^2}x = 0} \end{array} \Leftrightarrow \sin 2x = 0 \Leftrightarrow 2x = k\pi \Leftrightarrow x = \frac{{k\pi }}{2}} \right.} \right.} \right.\)