Bài 1:

1: \(y=\frac{\sin x+2\cdot cosx+1}{2\cdot\sin x+cosx+3}\)

=>\(2y\cdot\sin x+y\cdot cosx+3y=\sin x+2\cdot cosx+1\)

=>\(\left(2y-1\right)\cdot\sin x+cosx\cdot\left(y-2\right)=1-3y\)

Để phương trình có nghiệm thì \(\left(2y-1\right)^2+\left(y-2\right)^2>=\left(1-3y\right)^2\)

=>\(4y^2-4y+1+y^2-4y+4\ge9y^2-6y+1\)

=>\(5y^2-8y+5-9y^2+6y-1\ge0\)

=>\(-4y^2-2y+4\ge0\)

=>\(y^2+\frac12y-1\le0\)

=>\(y^2+2\cdot y\cdot\frac14+\frac{1}{16}-\frac{17}{16}\le0\)

=>\(\left(y+\frac14\right)^2\le\frac{17}{16}\)

=>\(-\frac{\sqrt{17}}{4}\le y+\frac14\le\frac{\sqrt{17}}{4}\)

=>\(\frac{-\sqrt{17}-1}{4}\le y\le\frac{\sqrt{17}-1}{4}\)

=>\(y_{\min}=\frac{-\sqrt{17}-1}{4}\) và \(y_{\max}=\frac{\sqrt{17}-1}{4}\)

2: \(y=2\cdot\sin^2x-3\cdot\sin x\cdot cosx+cos^2x\)

\(=2\cdot\frac{1-cos2x}{2}-3\cdot\frac12\cdot\sin2x+\frac{1+cos2x}{2}\)

\(=1-cos2x-\frac32\cdot\sin2x+\frac12+\frac12\cdot cos2x\)

\(=-\frac32\cdot\sin2x-\frac12\cdot cos2x+\frac32=-\frac12\left(3\cdot\sin2x+cos2x-3\right)\)

\(=-\frac{\sqrt{10}}{2}\left(\frac{3}{\sqrt{10}}\cdot\sin2x+\frac{1}{\sqrt{10}}\cdot cos2x-\frac{3}{\sqrt{10}}\right)\)

\(=-\frac{\sqrt{10}}{2}\cdot\left\lbrack\sin\left(2x+\alpha\right)-\frac{3}{\sqrt{10}}\right\rbrack\) , với \(cosa=\frac{3}{\sqrt{10}};\sin a=\frac{1}{\sqrt{10}}\)

\(=-\frac{\sqrt{10}}{2}\cdot\sin\left(2x+\alpha\right)+\frac32\)

Ta có: \(-1\le\sin\left(2x+a\right)\le1\)

=>\(-1\cdot\frac{-\sqrt{10}}{2}\ge\frac{-\sqrt{10}}{2}\sin\left(2x+a\right)\ge1\cdot\frac{-\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)\le\frac{\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}+\frac32\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)+\frac32\le\frac{\sqrt{10}}{2}+\frac32\)

=>\(y_{\min}=\frac{-\sqrt{10}+3}{2};y_{\max}=\frac{\sqrt{10}+3}{2}\)

Đáp án D

Bài 1: ĐKXĐ: cosx-1<>0

=>cosx<>1

=>\(x<>k2\pi\)

mà x∈[0;4pi]

nên x∉{0;2pi;4pi}

Ta có: \(\frac{\sin x}{cosx-1}=0\)

=>sin x=0

=>\(x=k\pi\)

mà \(x\in\left\lbrace0;4\pi\right\rbrace;x<>k2\pi\)

nên \(x\in\left\lbrace\pi;3\pi\right\rbrace\)

=>Tổng tất cả nghiệm là: \(\pi+3\pi=4\pi\)

Chọn D

Đáp án D.

Ta có

\(\left|cosx\right|-\left|sinx\right|-\left(\left|cosx\right|-\left|sinx\right|\right)\left(\left|cosx\right|+\left|sinx\right|\right)\sqrt{1+sin2x}=0\)

\(\Leftrightarrow\left(\left|cosx\right|-\left|sinx\right|\right)\left(1-\left(\left|cosx\right|+\left|sinx\right|\right)\sqrt{1+sin2x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|cosx\right|=\left|sinx\right|\Leftrightarrow cos2x=0\left(1\right)\\\left(\left|cosx\right|+\left|sinx\right|\right)\sqrt{1+sin2x}=1\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

\(\left(2\right)\Leftrightarrow\left|cosx\right|+\left|sinx\right|=\dfrac{1}{\sqrt{1+sin2x}}\) (với \(sin2x\ne-1\))

\(\Leftrightarrow1+2\left|sinx.cosx\right|=\dfrac{1}{1+sin2x}\)

\(\Leftrightarrow1+\left|sin2x\right|=\dfrac{1}{1+sin2x}\)

TH1: \(-1< sin2x< 0\Rightarrow1-sin2x=\dfrac{1}{1+sin2x}\)

\(\Leftrightarrow1-sin^22x=1\Rightarrow sin2x=0\) (loại)

TH2: \(0\le sin2x\le1\Rightarrow1+sin2x=\dfrac{1}{1+sin2x}\)

\(\Leftrightarrow1+sin2x=1\Leftrightarrow sin2x=0\Rightarrow x=\dfrac{k\pi}{2}\)

Vậy \(\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)

Bạn tự tìm số giá trị nhé

@Nguyễn Việt Lâm giúp em với

1: =>sin^2(3x)=0

=>sin 3x=0

=>3x=kpi

=>x=kpi/3

2:

\(sinx=1-cos^2x=sin^2x\)

=>\(sin^2x-sinx=0\)

=>sin x(sin x-1)=0

=>sin x=0 hoặc sin x=1

=>x=pi/2+k2pi hoặc x=kpi

4:

sin 2x+sin x=0

=>sin 2x=-sin x=sin(-x)

=>2x=-x+k2pi hoặc 2x=pi+x+k2pi

=>x=pi+k2pi hoặc x=k2pi/3

5: =>cos(x+pi/3)=1/căn 2

=>x+pi/3=pi/4+k2pi hoặc x+pi/3=-pi/4+k2pi

=>x=-pi/12+k2pi hoặc x=-7/12pi+k2pi