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\(\Leftrightarrow\sin^2x-\sin^22x+\sin^23x-\sin^24x=0\)
\(\Leftrightarrow\left(\sin x+\sin2x\right)\left(\sin2x-\sin x\right)+\left(\sin3x+\sin4x\right)\left(\sin4x-\sin3x\right)=0\)
\(\Leftrightarrow2\sin\dfrac{3}{2}x.\cos\dfrac{x}{2}.2\cos\dfrac{3}{2}x.\sin\dfrac{x}{2}+2\sin\dfrac{7}{2}x.\cos\dfrac{x}{2}.2\sin\dfrac{x}{2}\cos\dfrac{7}{2}x=0\)
\(\Leftrightarrow\sin3x.\sin x+\sin7x.\sin x=0\)
\(\Leftrightarrow\sin x\left(\sin3x+\sin7x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin x=0\\\sin3x=\sin\left(-7x\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\3x=-7x+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)
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Chứng minh các biểu thức đã cho không phụ thuộc vào x.
f(x) = 1 ⇒ f′(x) = 0
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\(pt\Leftrightarrow cos6x+3cos2x-4\left(2cos^2x-1\right)=0\)
\(\Leftrightarrow cos6x+3cos2x-4cos2x=0\)
\(\Leftrightarrow cos6x-cos2x=0\)
\(\Leftrightarrow-2sin4x.sin2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{k\pi}{4}\)
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\(y=\dfrac{4}{3}\left(\sin^6x+\cos^6x\right)+\cos4x-1\)
\(\sin^6x+\cos^6x=\left(\sin^2x+\cos^2x\right)\left(\sin^4x-\sin^2x\cdot\cos^2x+\cos^4x\right)\\ =\left(\sin^2x+\cos^2x\right)^2-3\sin^2x\cdot\cos^2x=1-\dfrac{3}{4}\sin^22x\)
Do \(0\le\sin^22x\le1\Leftrightarrow\dfrac{3}{4}\cdot0\ge-\dfrac{3}{4}\sin^22x\ge-\dfrac{3}{4}\)
\(\Leftrightarrow1\ge1-\dfrac{3}{4}\sin^22x\ge1-\dfrac{3}{4}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{4}{3}\ge\dfrac{4}{3}\left(\sin^6x+\cos^6x\right)\ge\dfrac{1}{4}\cdot\dfrac{4}{3}=\dfrac{1}{3}\)
Ta có \(-1\le\cos4x\le1\)
\(\Leftrightarrow\dfrac{1}{3}-1-1\le\dfrac{4}{3}\left(\sin^6x+\cos^6x\right)+\cos4x-1\le\dfrac{4}{3}+1-1\\ \Leftrightarrow-\dfrac{5}{3}\le y\le\dfrac{4}{3}\)
Vậy \(y_{min}=-\dfrac{5}{3};y_{max}=\dfrac{4}{3}\)
\(y=\dfrac{4}{3}\left(sin^6x+cos^6x\right)+cos4x-1\)
\(y=\dfrac{4}{3}\left(\dfrac{5}{8}+\dfrac{3}{8}cos4x\right)+cos4x-1\)
\(y=\dfrac{3}{2}cos4x-\dfrac{1}{6}\)
\(-1\le cos4x\le1\Rightarrow-\dfrac{5}{3}\le y\le\dfrac{4}{3}\)
\(y_{min}=-\dfrac{5}{3}\) khi \(cos4x=-1\)
\(y_{max}=\dfrac{4}{3}\) khi \(cos4x=1\)
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\(\Leftrightarrow1-cos6x-1-cos8x=1-cos10x-1-cos12x\)
\(\Leftrightarrow cos10x+cos12x-cos6x-cos8x=0\)
\(\Leftrightarrow cos11x.cosx+cos7x.cosx=0\)
\(\Leftrightarrow cosx\left(cos11x+cos7x\right)=0\)
\(\Leftrightarrow cosx.cos9x.cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=0\\cos9x=0\end{matrix}\right.\) \(\Leftrightarrow...\)
\(2\sqrt{3}\sin^23x-cos6x=\sqrt{3}\\ \Leftrightarrow2sin^23x\left(\sqrt{3}+1\right)=\sqrt{3}+1\\ \Leftrightarrow sin^23x=\dfrac{1}{2}\Leftrightarrow\dfrac{1-cos6x}{2}=\dfrac{1}{2}\Rightarrow cos6x=0\Leftrightarrow x=\pm\dfrac{\pi}{12}+k\dfrac{\pi}{6}\left(k\in Z\right)\)