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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 \(x=0\)
\(y=4\left(1-sin^2x\right)+2sinx+2=-4sin^2x+2sinx+6\)
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 \(y_{max}=\dfrac{97}{16}\)
2.
$y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$
$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$
Vì: $0\leq \sin ^22x\leq 1$
$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$
Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$
3.
$0\leq |\sin x|\leq 1$
$\Rightarrow 3\geq 3-2|\sin x|\geq 1$
Vậy $y_{\min}=1; y_{\max}=3$
`TXĐ: R`
Ta có: `-1 <= sin(x+ \pi/3) <= 1`
`<=>0 <= sin^4 (x+\pi/3) <= 1`
`<=>2 <= y <= 3`
`=>y_[mi n]=2<=>sin(x +\pi/3)=0<=>x= -\pi/3+k\pi` `(k in ZZ)`
`y_[max]=3<=>sin(x +\pi/3)=1<=>x=\pi/6 +k2\pi` `(k in ZZ)`
\(y=4cos^2\left(\dfrac{x}{2}-\dfrac{\pi}{12}\right)-7=2\left[cos\left(x-\dfrac{\pi}{6}\right)+1\right]-7=2cos\left(x-\dfrac{\pi}{6}\right)-5\)
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 \(x=\pi\)
24.
\(cos\left(x-\dfrac{\pi}{2}\right)\le1\Rightarrow y\le3.1+1=4\)
\(y_{max}=4\)
26.
\(y=\sqrt{2}cos\left(2x-\dfrac{\pi}{4}\right)\)
Do \(cos\left(2x-\dfrac{\pi}{4}\right)\le1\Rightarrow y\le\sqrt{2}\)
\(y_{max}=\sqrt{2}\)
b.
\(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)