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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 \(cos4x=1\)
Đặt \(sinx=t\left(t\in\left[-1;1\right]\right)\).
\(\Rightarrow y=f\left(t\right)=-2t^2+3t-1\)
\(\Rightarrow y_{min}=min\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{3}{4}\right)\right\}=f\left(-1\right)=-6\)
\(y_{max}=max\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{3}{4}\right)\right\}=f\left(\dfrac{3}{4}\right)=\dfrac{1}{8}\)
\(y=\sqrt{3}cos2x+2sinxcosx-2\)
\(=\sqrt{3}cos2x+sin2x-2\)
Ta có: \(\left|\sqrt{3}cos2x+sin2x\right|\le\sqrt{\left(\sqrt{3}\right)^2+1^2}=2\)
Do đó \(-2\le\sqrt{3}cos2x+sin2x\le2\)
\(\Leftrightarrow-4\le\sqrt{3}cos2x+sin2x-2\le2\).
Ta có: \(\left|\sqrt{3}cosx-sinx\right|\le\sqrt{\left(\sqrt{3}\right)^2+\left(-1\right)^2}=2\)
Do đó \(-2\le\sqrt{3}cosx-sinx\le2\)
a.
\(y=sinx.cosx+1=\dfrac{1}{2}sin2x+1\)
\(-1\le sin2x\le1\Rightarrow\dfrac{1}{2}\le y\le\dfrac{3}{2}\)
\(y_{min}=\dfrac{1}{2}\) khi \(sin2x=-1\Rightarrow x=-\dfrac{\pi}{4}+k\pi\)
\(y_{max}=\dfrac{3}{2}\) khi \(sin2x=1\Rightarrow x=\dfrac{\pi}{4}+k\pi\)
b.
\(y=2\left(\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx\right)-2=2.sin\left(x-\dfrac{\pi}{6}\right)-2\)
\(-1\le sin\left(x-\dfrac{\pi}{6}\right)\le1\Rightarrow-4\le y\le0\)
\(y_{min}=-4\) khi \(sin\left(x-\dfrac{\pi}{6}\right)=-1\Rightarrow x=-\dfrac{\pi}{3}+k2\pi\)
\(y_{max}=0\) khi \(sin\left(x-\dfrac{\pi}{6}\right)=1\Rightarrow x=\dfrac{2\pi}{3}+k2\pi\)
\(y=\sqrt{3}cosx-sinx=2\left(\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx\right)=2cos\left(x+\dfrac{\pi}{6}\right)\)
Vì \(cos\left(x+\dfrac{\pi}{6}\right)\in\left[-1;1\right]\Rightarrow y=\sqrt{3}cosx-sinx\in\left[-2;2\right]\)
\(\Rightarrow y_{min}=-2\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=-1\Leftrightarrow x+\dfrac{\pi}{6}=\pi+k2\pi\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\)
\(y_{max}=2\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=1\Leftrightarrow x+\dfrac{\pi}{6}=k2\pi\Leftrightarrow x=-\dfrac{\pi}{6}+k2\pi\)
