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\(y=2sin^2x+3sinx.cosx+cos^2x\)
\(=-\left(1-2sin^2x\right)+\dfrac{3}{2}sin2x+\dfrac{1}{2}\left(2cos^2x-1\right)+\dfrac{1}{2}\)
\(=-cos2x+\dfrac{3}{2}sin2x+\dfrac{1}{2}cos2x+\dfrac{1}{2}\)
\(=\dfrac{3}{2}sin2x-\dfrac{1}{2}cos2x+\dfrac{1}{2}\)
\(=\dfrac{\sqrt{10}}{2}\left(\dfrac{3}{\sqrt{10}}sin2x-\dfrac{1}{\sqrt{10}}cos2x\right)+\dfrac{1}{2}\)
\(=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\)
Vì \(sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)\in\left[-1;1\right]\)
\(\Rightarrow y=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\in\left[-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2};\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\right]\)
\(\Rightarrow y_{min}=-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=-1\Leftrightarrow...\)
\(y_{max}=\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=1\Leftrightarrow...\)
Tham khảo: tìm GTLN - GTNN của hàm số : y=sinx cosx sinxcosx - Hoc24
Đặt
Xét hàm
Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=t\Rightarrow t\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)
\(\Rightarrow y=t+\dfrac{t^2-1}{2}=\dfrac{1}{2}t^2+t-\dfrac{1}{2}\)
Xét hàm \(y=f\left(t\right)=\dfrac{1}{2}t^2+t-\dfrac{1}{2}\) trên \(\left[-\sqrt{2};\sqrt{2}\right]\)
\(-\dfrac{b}{2a}=-1\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(f\left(-\sqrt{2}\right)=\dfrac{1-2\sqrt{2}}{2}\) ; \(f\left(-1\right)=-1\) ; \(f\left(\sqrt{2}\right)=\dfrac{1+2\sqrt{2}}{2}\)
\(\Rightarrow y_{min}=-1\) ; \(y_{max}=\dfrac{1+2\sqrt{2}}{2}\)
\(y=sin\left(x+\dfrac{\pi}{3}\right)-sinx\)
\(=\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx-sinx\)
\(=\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx\)
\(=cos\left(x+\dfrac{\pi}{6}\right)\in\left[-1;1\right]\)
\(\Rightarrow\left\{{}\begin{matrix}y_{mịn}=-1\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\\y_{max}=1\Leftrightarrow x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(y=\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x+\frac{1}{2}=sin\left(2x+\frac{\pi}{6}\right)+\frac{1}{2}\)
Do \(-1\le sin\left(2x+\frac{\pi}{6}\right)\le1\Rightarrow-\frac{1}{2}\le y\le\frac{3}{2}\)
\(y_{min}=-\frac{1}{2}\) khi \(sin\left(2x+\frac{\pi}{6}\right)=-1\)
\(y_{max}=\frac{3}{2}\) khi \(sin\left(2x+\frac{\pi}{6}\right)=1\)
a, Đặt \(t=cos3x\left(t\in\left[-1;1\right]\right)\)
\(y=9-sin^23x-\sqrt{2}cos3x\)
\(=cos^23x-\sqrt{2}cos3x+8\)
\(\Leftrightarrow y=f\left(t\right)=t^2-\sqrt{2}t+8\)
\(\Rightarrow minf\left(t\right)\le y\le maxf\left(x\right)\)
\(\Rightarrow min\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{\sqrt{2}}{2}\right)\right\}\le y\le max\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{\sqrt{2}}{2}\right)\right\}\)
\(\Rightarrow\dfrac{15}{2}\le y\le9+\sqrt{2}\)
\(\Rightarrow y_{max}=9+\sqrt{2}\)
b, Đặt \(t=sin3x\left(t\in\left[-1;1\right]\right)\)
\(y=3sin3x-8cos^23x+4\)
\(=3sin3x+8-8cos^23x-4\)
\(=8sin^23x+3sin3x-4\)
\(\Leftrightarrow y=f\left(t\right)=8t^2+3t-4\)
\(\Rightarrow minf\left(x\right)\le y\le maxf\left(t\right)\)
\(\Rightarrow min\left\{f\left(-1\right);f\left(1\right);f\left(-\dfrac{3}{16}\right)\right\}\le y\le max\left\{f\left(-1\right);f\left(1\right);f\left(-\dfrac{3}{16}\right)\right\}\)
\(\Rightarrow-\dfrac{137}{32}\le y\le7\)
\(\Rightarrow y_{max}=7\)
ĐKXĐ: \(sinx;cosx\ge0\)
Do \(\left\{{}\begin{matrix}0\le sinx\le1\\0\le cosx\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{sinx}\ge sin^2x\\\sqrt{cosx}\ge cos^2x\end{matrix}\right.\)
\(\Rightarrow\sqrt{sinx}+\sqrt{cosx}\ge sin^2x+cos^2x=1\)
\(\Rightarrow y_{min}=1\) (khi \(x=\dfrac{\pi}{2}+k2\pi\) hoặc \(k2\pi\))
Mặt khác áp dụng Bunhiacopxki:
\(y\le\sqrt{2\left(sinx+cosx\right)}\le\sqrt{2\sqrt{2\left(sin^2x+cos^2x\right)}}=\sqrt[4]{8}\)
\(y_{max}=\sqrt[4]{8}\) khi \(x=\dfrac{\pi}{4}+k2\pi\)
ĐK: Biểu thức xác định với mọi `x`.
`y_(min) <=> (\sqrt(2-cos(x-π/6))+3)_(max) <=> (cos(x-π/6))_(max)`
`<=> cos(x-π/6)=1 <=> x-π/6=k2π <=> x = π/6+k2π ( k \in ZZ)`.
`=> y_(min) = 1`
`y_(max) <=> (\sqrt(2-cos(x-π/6))+3)_(min) <=> (cos(x-π/6))_(min)`
`<=> cos(x-π/6)=-1 <=> x -π/6= π+k2π <=> x = (7π)/6+k2π (k \in ZZ)`
`=> y_(max) = (6-2\sqrt3)/3`.