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 saiĐK: \(4 \cos ^{2} 2 x+\sin ^{2} 2 x \neq 0 \Leftrightarrow 4 \cos ^{2} 2 x+1-\cos ^{2} 2 x \neq 0 \Leftrightarrow 3 \cos ^{2} 2 x+1 \neq 0 \Leftrightarrow \forall x \in \mathbb{R}\)
\(\begin{aligned} &P T \Leftrightarrow \frac{\sin ^{10} x+\cos ^{10} x}{4}=\frac{\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)}{4\left(1-\sin ^{2} 2 x\right)+\sin ^{2} 2 x}\\ &\Leftrightarrow \frac{\sin ^{10} x+\cos ^{10} x}{4}=\frac{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-3 \sin ^{2} x \cos ^{2} x}{4-3 \sin ^{2} 2 x}\\ &\Leftrightarrow \frac{\sin ^{10} x+\cos ^{10} x}{4}=\frac{1-\frac{3}{4} \sin ^{2} 2 x}{4-3 \sin ^{2} 2 x} \Leftrightarrow \frac{\sin ^{10} x+\cos ^{10} x}{4}=\frac{4-3 \sin ^{2} 2 x}{4\left(4-3 \sin ^{2} 2 x\right)}\\ &\Leftrightarrow \sin ^{10} x+\cos ^{10} x=1 \Leftrightarrow \sin ^{10} x+\cos ^{10} x=\sin ^{2} x+\cos ^{2} x\\ &\Leftrightarrow \sin ^{2} x\left(1-\sin ^{8} x\right)+\cos ^{2} x\left(1-\cos ^{8} x\right)=0 \end{aligned}\,\,\,(*)\)
Vì \(\left\{\begin{array}{l} \sin ^{2} x\left(1-\sin ^{8} x\right) \geq 0 \forall x \in \mathbb{R} \\ \cos ^{2} x\left(1-\cos ^{8} x\right) \geq 0 \forall x \in \mathbb{R} \end{array}\right. \text { nên }(*) \Leftrightarrow\left\{\begin{array}{l} \sin ^{2} x\left(1-\sin ^{8} x\right)=0 \\ \cos ^{2} x\left(1-\cos ^{8} x\right)=0 \end{array}\right.\)
\( \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} \sin x = 0\\ \sin x = \pm 1 \end{array} \right.\\ \left[ \begin{array}{l} \cos x = 0\\ \cos x = \pm 1 \end{array} \right. \end{array} \right. \Leftrightarrow \Leftrightarrow x = \frac{{k\pi }}{2}\)
