K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

1: \(y=\sqrt{3}\cdot sin^2x-\left(1-sin^2x\right)+5\)

\(=sin^2x\left(\sqrt{3}+1\right)-1+5=sin^2x\left(\sqrt{3}+1\right)+4\)

\(0< =sin^2x< =1\)

=>\(0< =sin^2x\left(\sqrt{3}+1\right)< =\sqrt{3}+1\)

=>4<=y<=căn 3+5

y min=4 khi sin^2x=0

=>sin x=0

=>x=kpi

\(y_{max}=5+\sqrt{3}\) khi \(sin^2x=1\)

=>\(cos^2x=0\)

=>cosx=0

=>\(x=\dfrac{pi}{2}+kpi\)

2: \(y=5\left[\dfrac{3}{5}sinx+\dfrac{4}{5}cosx\right]+7\)

\(=5\cdot\left[sinx\cdot cosa+cosx\cdot sina\right]+7\)(Với cosa=3/5; sin a=4/5)

\(=5\cdot sin\left(x+a\right)+7\)

-1<=sin(x+a)<=1

=>-5<=5sin(x+a)<=5

=>-5+7<=y<=5+7

=>2<=y<=12

\(y_{min}=2\) khi sin (x+a)=-1

=>x+a=-pi/2+kp2i

=>\(x=-\dfrac{pi}{2}+k2pi-a\)

\(y_{max}=12\) khi sin(x+a)=1

=>x+a=pi/2+k2pi

=>\(x=\dfrac{pi}{2}+k2pi-a\)

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 :?)

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/ \(3+1\ge y^2\Leftrightarrow4\ge y^2\Leftrightarrow-2\le y\le2\)

\(y_{max}=2\Leftrightarrow\sqrt{3}\sin x+\cos x=2\Leftrightarrow\dfrac{\sqrt{3}}{2}\sin x+\dfrac{1}{2}\cos x=1\Leftrightarrow\cos\dfrac{\pi}{6}.\sin x+\sin\dfrac{\pi}{6}.\cos x=1\)

\(\Rightarrow\sin\left(x+\dfrac{\pi}{6}\right)=1\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\Leftrightarrow x=\dfrac{\pi}{3}+k2\pi\)

\(y_{min}=-2\Leftrightarrow\sin\left(x+\dfrac{\pi}{6}\right)=-1\Leftrightarrow x+\dfrac{\pi}{6}=-\dfrac{\pi}{2}+k2\pi\Leftrightarrow x=-\dfrac{2}{3}\pi+k2\pi\)

16 tháng 8 2021

a) <=> 4sinxcosx -(2cos2x-1)=7sinx+2cosx-4

<=> 2cos2x+(2-4sinx)cosx+7sinx-5=0

- sinx=1 => 2cos2x-2cosx+2=0 

pt trên vn

16 tháng 8 2021

b) <=> 2sinxcosx-1+2sin2x+3sinx-cosx-1=0

<=> cos(2sinx-1)+2sin2x+3sinx-2=0

<=> cosx(2sinx-1)+(2sinx-1)(sinx+2)=0

<=> (2sinx-1)(cosx+sinx+2)=0

<=> sinx=1/2 hoặc cosx+sinx=-2(vn)

<=> x= \(\frac{\pi}{6}+k2\pi\) hoặc \(x=\frac{5\pi}{6}+k2\pi\left(k\in Z\right)\)

NV
20 tháng 9 2021

\(y=2\left(\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x\right)=2sin\left(2x+\dfrac{\pi}{3}\right)\)

\(-1\le sin\left(2x+\dfrac{\pi}{3}\right)\le1\Rightarrow-2\le y\le2\)

\(y_{min}=-2\) khi \(sin\left(2x+\dfrac{\pi}{3}\right)=-1\Rightarrow x=-\dfrac{5\pi}{12}+k\pi\)

\(y_{max}=2\) khi \(sin\left(2x+\dfrac{\pi}{3}\right)=1\Rightarrow x=\dfrac{\pi}{12}+k\pi\)

a: \(y=\sqrt{2}sin\left(x+\dfrac{pi}{4}\right)\)

\(-1< =sin\left(x+\dfrac{pi}{4}\right)< =1\)

=>\(-\sqrt{2}< =y< =\sqrt{2}\)

\(y_{min}=-\sqrt{2}\) khi sin(x+pi/4)=-1

=>x+pi/4=-pi/2+k2pi

=>x=-3/4pi+k2pi

\(y_{max}=\sqrt{2}\) khi sin(x+pi/4)=1

=>x+pi/4=pi/2+k2pi

=>x=pi/4+k2pi

b: \(y=sinx\cdot cos\left(\dfrac{pi}{3}\right)+cosx\cdot sin\left(\dfrac{pi}{3}\right)+3\)

\(=sin\left(x+\dfrac{pi}{3}\right)+3\)

-1<=sin(x+pi/3)<=1

=>-1+3<=sin(x+pi/3)+3<=4

=>2<=y<=4

y min=2 khi sin(x+pi/3)=-1

=>x+pi/3=-pi/2+k2pi

=>x=-5/6pi+k2pi

y max=4 khi sin(x+pi/3)=1

=>x+pi/3=pi/2+k2pi

=>x=pi/6+k2pi

c: \(y=2\cdot\left(sin2x\cdot\dfrac{\sqrt{3}}{2}-cos2x\cdot\dfrac{1}{2}\right)\)

\(=2sin\left(2x-\dfrac{pi}{6}\right)\)

-1<=sin(2x-pi/6)<=1

=>-2<=y<=2

y min=-2 khi sin(2x-pi/6)=-1

=>2x-pi/6=-pi/2+k2pi

=>2x=-1/3pi+k2pi

=>x=-1/6pi+kpi

y max=2 khi sin(2x-pi/6)=1

=>2x-pi/6=pi/2+k2pi

=>2x=2/3pi+k2pi

=>x=1/3pi+kpi

7 tháng 3 2017

NV
16 tháng 7 2021

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.

\(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

2 tháng 11 2020

Xét phương trình: y=3sinx+4cosx+5

<=>3sinx+4cosx+5-y=0

Để phương trình có nghiệm:

=>32+42≥(5-y)2 (đẳng thức Bunhiacopxki)

<=>25≥25-10y+y2

<=>y2-10y≤0

<=>0≤y≤10

vậy miny=0; maxy=10