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\(y=4\left(1-sin^2x\right)+2sinx+2=-4sin^2x+2sinx+6\)
Đặt \(sinx=t\in\left[-1;1\right]\Rightarrow y=f\left(t\right)=-4t^2+2t+6\)
\(-\dfrac{b}{2a}=\dfrac{1}{4}\in\left[-1;1\right]\)
\(f\left(-1\right)=0\) ; \(f\left(\dfrac{1}{4}\right)=\dfrac{25}{4}\); \(f\left(1\right)=4\)
\(\Rightarrow y_{max}=\dfrac{25}{4}\) khi \(sinx=\dfrac{1}{4}\)
\(y_{min}=0\) khi \(sinx=-1\)
Ta có: \(y=4cos^2x+2sinx+2=4-4sin^2x+2sinx+2=-4sin^2x+2sinx+6=-\left(4sin^2x-2sinx+\dfrac{1}{16}-\dfrac{1}{16}-6\right)=-\left(2sin^2x-\dfrac{1}{4}\right)^2+\dfrac{97}{16}\)
Ta có: \(-\left(2sin^2x-\dfrac{1}{4}\right)^2\le0\Rightarrow y\le\dfrac{97}{16}\)
Vậy \(y_{max}=\dfrac{97}{16}\)
Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\Rightarrow\left\{{}\begin{matrix}-\sqrt{2}\le t\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)
\(\Rightarrow y=t+t^2-1-1=t^2+t-2\)
Xét hàm \(f\left(t\right)=t^2+t-2\) trên \(\left[-\sqrt{2};\sqrt{2}\right]\)
\(-\frac{b}{2a}=-\frac{1}{2}\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(f\left(-\frac{1}{2}\right)=-\frac{9}{4}\) ; \(f\left(-\sqrt{2}\right)=-\sqrt{2}\) ; \(f\left(\sqrt{2}\right)=\sqrt{2}\)
\(\Rightarrow y_{max}=\sqrt{2}\) khi \(t=\sqrt{2}\)
\(y_{min}=-\frac{9}{4}\) khi \(t=-\frac{1}{2}\)
\(y=4cos^2\left(\dfrac{x}{2}-\dfrac{\pi}{12}\right)-7=2\left[cos\left(x-\dfrac{\pi}{6}\right)+1\right]-7=2cos\left(x-\dfrac{\pi}{6}\right)-5\)
Đặt \(x-\dfrac{\pi}{6}=t\Rightarrow t\in\left[-\dfrac{\pi}{6};\dfrac{5\pi}{6}\right]\)
\(\Rightarrow y=2cost-5\)
Do \(t\in\left[-\dfrac{\pi}{6};\dfrac{5\pi}{6}\right]\Rightarrow cost\in\left[-\dfrac{\sqrt{3}}{2};1\right]\)
\(\Rightarrow y\in\left[-5-\sqrt{3};-3\right]\)
\(y_{max}=-3\) khi \(t=0\) hay \(x=\dfrac{\pi}{6}\)
\(y_{min}=-5-\sqrt{3}\) khi \(y=\dfrac{5\pi}{6}\) hay \(x=\pi\)
\(y=\frac{2cos2x+2+3sin2x+1}{3-sin2x+cos2x}=\frac{2cos2x+3sin2x+3}{3-sin2x+cos2x}\)
\(\Leftrightarrow3y-y.sin2x+y.cos2x=2cos2x+3sin2x+3\)
\(\Leftrightarrow\left(y+3\right)sin2x+\left(2-y\right)cos2x=3y-3\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(\left(y+3\right)^2+\left(2-y\right)^2\ge\left(3y-3\right)^2\)
\(\Leftrightarrow7y^2-20y-4\le0\)
\(\Leftrightarrow\frac{10-8\sqrt{2}}{7}\le y\le\frac{10+8\sqrt{2}}{7}\)
\(\Rightarrow\left\{{}\begin{matrix}M=\frac{10+8\sqrt{2}}{7}\\m=\frac{10-8\sqrt{2}}{7}\end{matrix}\right.\) \(\Rightarrow7M-14m=24\sqrt{2}-10\)
a.
\(-1\le sin\left(1-x^2\right)\le1\)
\(\Rightarrow y_{min}=-1\) khi \(1-x^2=-\dfrac{\pi}{2}+k2\pi\) \(\Rightarrow x^2=\dfrac{\pi}{2}+1+k2\pi\) (\(k\ge0\))
\(y_{max}=1\) khi \(1-x^2=\dfrac{\pi}{2}+k2\pi\Rightarrow x^2=1-\dfrac{\pi}{2}+k2\pi\) (\(k\ge1\))
b.
Đặt \(\sqrt{2-x^2}=t\Rightarrow t\in\left[0;\sqrt{2}\right]\subset\left[0;\pi\right]\)
\(y=cost\) nghịch biến trên \(\left[0;\pi\right]\Rightarrow\) nghịch biến trên \(\left[0;\sqrt{2}\right]\)
\(\Rightarrow y_{max}=y\left(0\right)=cos0=1\) khi \(x^2=2\Rightarrow x=\pm\sqrt{2}\)
\(y_{min}=y\left(\sqrt{2}\right)=cos\sqrt{2}\) khi \(x=0\)
