tìm GTLN,GTNN của hàm số
y= 2sin4x+3
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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 \(cos4x=1\)
b: \(y=\sin x+\sqrt3\cdot cosx\)
\(=2\left(\frac12\cdot\sin x+\frac{\sqrt3}{2}\cdot cosx\right)\)
\(=2\cdot\sin\left(x+\frac{\pi}{3}\right)\)
TA có: \(-1\le\sin\left(x+\frac{\pi}{3}\right)\le1\)
=>\(-2\le2\cdot\sin\left(x+\frac{\pi}{3}\right)\le2\)
=>-2<=y<=2
y min=-2 khi \(\sin\left(x+\frac{\pi}{3}\right)=-1\)
=>\(x+\frac{\pi}{3}=-\frac{\pi}{2}+k2\pi\)
=>\(x=-\frac56\pi+k2\pi\)
y max=2 khi \(\sin\left(x+\frac{\pi}{3}\right)=1\)
=>\(x+\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)
=>\(x=\frac{\pi}{2}-\frac{\pi}{3}+k2\pi=\frac{\pi}{6}+k2\pi\)
c: 0<=|cosx|<=1
=>0>=-2|cosx|>=-2
=>0+3>=-2|cosx|+3>=-2+3
=>3>=y>=1
y max=3 khi cosx=0
=>\(x=\frac{\pi}{2}+k\pi\)
y min=1 khi |cosx|=1
=>\(cos^2x=1\)
=>\(\sin^2x=0\)
=>sin x=0
=>\(x=k\pi\)
21.
a) `2sin(x-30^@)-1=0`
`<=>sin(x-30^@)=1/2`
`<=> sin(x-30^@)=sin30^@`
`<=>[(x-30^@=30^@+k360^@),(x-30^@=180^@-30^@+k360^@):}`
`<=> [(x=60^@+k360^@),(x=180^@+k360^@):}`
b) `5sin^2x+3cosx+3=0`
`<=>5(1-cos^2x)+3cosx+3=0`
`<=>-5cos^2x+3cosx+8=0`
`<=>(cosx+1)(cosx=8/5)=0`
`<=>[(cosx=-1),(cosx=8/5\ (VN)):}`
`<=>x=180^@+k360^@`
22.
`-1<=sin2x<=1`
`<=>2<=3+sin2x<=4`
`=> y_(min)=2 ; y_(max)=4`
y=(sin2x-3)^2-6
-1<=sin2x<=1
=>-4<=sin2x-3<=-2
=>4<=(sin2x-3)^2<=16
=>-2<=y<=10
y min khi sin2x-3=-2
=>sin 2x=1
=>2x=pi/2+k2pi
=>x=pi/4+kpi
y max khi sin 2x-3=-4
=>sin 2x=-1
=>2x=-pi/2+k2pi
=>x=-pi/4+kpi
ta có -1\(\le sin4x\le1\)
=> ymin=2.-1+3=1
GTNN y=1 khi x\(=\frac{-\pi}{8}+\frac{k2\pi}{ }\)với k thuộc Z
=> ymax= 2.1+3=5
GTLN y=5 khi x=\(\frac{\pi}{8}+\frac{k\pi}{2}\) với k thuộc Z
krubshsn