Tìm min, max
a) \(y=\sqrt{3}sinx-cosx+5\)
b) \(y=cosx+cos\left(x+\frac{\pi}{3}\right)\)
c) \(y=sinx+\sqrt{3}cosx+12\)
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14 tháng 8 2023
a: ĐKXĐ: 2*sin x+1<>0
=>sin x<>-1/2
=>x<>-pi/6+k2pi và x<>7/6pi+k2pi
b: ĐKXĐ: \(\dfrac{1+cosx}{2-cosx}>=0\)
mà 1+cosx>=0
nên 2-cosx>=0
=>cosx<=2(luôn đúng)
c ĐKXĐ: tan x>0
=>kpi<x<pi/2+kpi
d: ĐKXĐ: \(2\cdot cos\left(x-\dfrac{pi}{4}\right)-1< >0\)
=>cos(x-pi/4)<>1/2
=>x-pi/4<>pi/3+k2pi và x-pi/4<>-pi/3+k2pi
=>x<>7/12pi+k2pi và x<>-pi/12+k2pi
e: ĐKXĐ: x-pi/3<>pi/2+kpi và x+pi/4<>kpi
=>x<>5/6pi+kpi và x<>kpi-pi/4
f: ĐKXĐ: cos^2x-sin^2x<>0
=>cos2x<>0
=>2x<>pi/2+kpi
=>x<>pi/4+kpi/2
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2020
c/
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=cos3x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos3x\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=3x+k2\pi\\x+\frac{\pi}{3}=-3x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{12}+\frac{k\pi}{2}\end{matrix}\right.\)
d/
\(\Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=sin2x\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{3}=2x+k2\pi\\3x-\frac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{15}+\frac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2020
a/
\(\Leftrightarrow\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{6}\right)\)
\(\Rightarrow x+\frac{\pi}{3}=\pi-x-\frac{\pi}{6}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{4}+k\pi\)
b/
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=sin\frac{\pi}{12}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)=sin\frac{\pi}{12}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=\frac{\pi}{12}+k2\pi\\x+\frac{\pi}{6}=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k2\pi\\x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
21 tháng 5 2021
a)\(y=\sqrt{3}sinx+cosx=2\left(\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\right)\)\(=2\left(sinx.cos\dfrac{\pi}{6}+cosx.sin\dfrac{\pi}{6}\right)\)\(=2sin\left(x+\dfrac{\pi}{6}\right)\)
Có \(-1\le sin\left(x+\dfrac{\pi}{6}\right)\le1\) \(\Leftrightarrow-2\le2sin\left(x+\dfrac{\pi}{6}\right)\le2\)
\(\Leftrightarrow-2\le y\le2\)
miny=-2 \(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=-1\) \(\Leftrightarrow x+\dfrac{\pi}{6}=-\dfrac{\pi}{2}+2k\pi\left(k\in Z\right)\) \(\Leftrightarrow x=-\dfrac{2\pi}{3}+k2\pi\left(k\in Z\right)\)
maxy=2\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=1\) \(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)\(\Leftrightarrow x=\dfrac{\pi}{3}+k2\pi\left(k\in Z\right)\)
b) \(y=sin2x-cos2x=\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)\)
Có \(\sqrt{2}\ge\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)\ge-\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}\ge y\ge-\sqrt{2}\)
miny=\(-\sqrt{2}\) \(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=-1\)\(\Leftrightarrow2x-\dfrac{\pi}{4}=-\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)\(\Leftrightarrow x=-\dfrac{\pi}{8}+k\pi\left(k\in Z\right)\)
maxy=\(\sqrt{2}\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=1\)\(\Leftrightarrow x=\dfrac{3\pi}{8}+k\pi\left(k\in Z\right)\)
c) \(y=3sinx+4cosx=5\left(\dfrac{3}{5}sinx+\dfrac{4}{5}cosx\right)\)
Đặt \(cosa=\dfrac{3}{5}\) và \(sina=\dfrac{4}{5}\)(vì cos2a+sin2a=1)
\(y=5\left(sinx.cosa+cosx.sina\right)\)\(=5sin\left(x+a\right)\)
\(\Rightarrow-5\le y\le5\)
miny=-5 <=> \(sin\left(x+a\right)=-1\)\(\Leftrightarrow x=-\dfrac{\pi}{2}-arc.sina+k2\pi\left(k\in Z\right)\)
maxy=5 <=> \(sin\left(x+a\right)=1\)\(\Leftrightarrow x=\dfrac{\pi}{2}-arc.sina+k2\pi\left(k\in Z\right)\)
(P/s1:cái x ở câu c ấy trông nó ngu ngu??
P/s2:sau khi load lại câu hỏi ở 1 tab khác ,thấy 1 câu trả lời nhưng vẫn đăng vì cảm thấy bỏ đi hơi phí :?)
HT
21 tháng 5 2021
Áp dụng quy tắc sau: Nếu \(a\sin x+b\cos y=c\Leftrightarrow a^2+b^2\ge c^2\)
a/ \(3+1\ge y^2\Leftrightarrow4\ge y^2\Leftrightarrow-2\le y\le2\)
\(y_{max}=2\Leftrightarrow\sqrt{3}\sin x+\cos x=2\Leftrightarrow\dfrac{\sqrt{3}}{2}\sin x+\dfrac{1}{2}\cos x=1\Leftrightarrow\cos\dfrac{\pi}{6}.\sin x+\sin\dfrac{\pi}{6}.\cos x=1\)
\(\Rightarrow\sin\left(x+\dfrac{\pi}{6}\right)=1\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\Leftrightarrow x=\dfrac{\pi}{3}+k2\pi\)
\(y_{min}=-2\Leftrightarrow\sin\left(x+\dfrac{\pi}{6}\right)=-1\Leftrightarrow x+\dfrac{\pi}{6}=-\dfrac{\pi}{2}+k2\pi\Leftrightarrow x=-\dfrac{2}{3}\pi+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 4 2019
\(P=\frac{sin^2x+cos^2x+2sinx.cosx-1}{\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right).cotx}-\frac{1}{cosx-sinx}\)
\(=\frac{2sinx.cosx}{\left(cosx-sinx\right).\frac{cosx}{sinx}}-\frac{1}{cosx-sinx}=\frac{2sin^2x}{cosx-sinx}-\frac{1}{cosx-sinx}\)
\(=\frac{2sin^2x-1}{cosx-sinx}=\frac{2sin^2x-\left(sin^2x+cos^2x\right)}{cosx-sinx}=\frac{sin^2x-cos^2x}{cosx-sinx}\)
\(=\frac{\left(sinx-cosx\right)\left(sinx+cosx\right)}{cosx-sinx}=-\left(sinx+cosx\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2020
d/
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\frac{sin\left(3x-x\right)}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sin2x}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{2sinx.cosx}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\)
\(\Leftrightarrow tanx=\sqrt{3}\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2020
c/
ĐKXĐ: \(sin2x\ne0\)
\(\Leftrightarrow\frac{\frac{sinx}{cosx}-sinx}{sin^3x}=\frac{1}{cosx}\)
\(\Leftrightarrow sinx-sinx.cosx=sin^3x\)
\(\Leftrightarrow1-cosx=sin^2x\)
\(\Leftrightarrow1-cosx=1-cos^2x\)
\(\Leftrightarrow cos^2x-cosx=0\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\)
a/ \(y=2\left(\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx\right)+5=2sin\left(x-\frac{\pi}{6}\right)+5\)
Do \(-1\le sin\left(x-\frac{\pi}{6}\right)\le1\Rightarrow3\le y\le7\)
b/ \(y=2cos\left(x+\frac{\pi}{6}\right)cos\left(-\frac{\pi}{6}\right)=\sqrt{3}cos\left(x+\frac{\pi}{6}\right)\)
Do \(-1\le cos\left(x+\frac{\pi}{6}\right)\le1\Rightarrow-\sqrt{3}\le y\le\sqrt{3}\)
c/ \(y=2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)+12=2sin\left(x+\frac{\pi}{3}\right)+12\)
Do \(-1\le sin\left(x+\frac{\pi}{3}\right)\le1\Rightarrow10\le y\le14\)