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

Thấy : \(sinx+cosx+2\ge-1-1+2=0\)  . " = " ko xảy ra nên : \(sinx+cosx+2>0\) 

Suy ra : \(\left(y-1\right)sinx+\left(y-2\right)cosx=1-2y\)  (*)

(*) có no \(\Leftrightarrow\left(y-1\right)^2+\left(y-2\right)^2\ge\left(1-2y\right)^2\Leftrightarrow2y^2-6y+5\ge4y^2-4y+1\Leftrightarrow-2y^2-2y+4\ge0\)

\(\Leftrightarrow-y^2-y+2\ge0\)  \(\Leftrightarrow-2\le y\le1\)

Suy ra : Max y = 1 . Chọn B 

21 : \(cosx-\sqrt{3}sinx=0\) 

cos x = 0 thay vào : sin x = 0 ( L ) 

cos x khác 0 \(\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\); ta có : \(1-\sqrt{3}tanx=0\Leftrightarrow tanx=\dfrac{1}{\sqrt{3}}\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\left(k\in Z\right)\)

49.

\(\Leftrightarrow m.sin2x+2\left(cos2x+1\right)=m+5\)

\(\Leftrightarrow m.sin2x+2cos2x=m+3\)

Pt có nghiệm khi:

\(m^2+2^2\ge\left(m+3\right)^2\)

\(\Leftrightarrow6m\le-5\Rightarrow m\le-\dfrac{5}{6}\)

\(\Rightarrow m=\left\{-3;-2;-1\right\}\)

50.

\(\Leftrightarrow m.2sin^2x+4sinx.cosx+3m.2cos^2x=2\)

\(\Leftrightarrow m\left(1-cos2x\right)+2sin2x+3m\left(1+cos2x\right)=2\)

\(\Leftrightarrow m.cos2x+sin2x=1-2m\)

Pt có nghiệm khi:

\(m^2+1\ge\left(1-2m\right)^2\Leftrightarrow3m^2-4m\le0\)

\(\Rightarrow m\in\left[0;\dfrac{4}{3}\right]\)

51.

ĐKXĐ: \(x\ne k\pi\)

\(\dfrac{5-4cosx}{sinx}=\dfrac{6tana}{1+tan^2a}\)

\(\Leftrightarrow\dfrac{5-4cosx}{sinx}=\dfrac{6sina}{cosa}.cos^2a=3sin2a\)

\(\Leftrightarrow5-4cosx=3sin2a.sinx\)

\(\Leftrightarrow3sin2a.sinx+4cosx=5\)

Pt có nghiệm khi:

\(\left(3sin2a\right)^2+4^2\ge5^2\)

\(\Leftrightarrow sin^22a\ge1\)

\(\Leftrightarrow sin^22a=1\Leftrightarrow cos2a=0\)

\(\Leftrightarrow2a=\dfrac{\pi}{2}+k\pi\Rightarrow a=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

\(\Rightarrow a=\left\{\dfrac{\pi}{4};\dfrac{3\pi}{4};\dfrac{5\pi}{4};\dfrac{7\pi}{4}\right\}\)

Em tự cộng và chọn kết quả nhé

13.

\(y=1+sin2x-\left(1-sin^22x\right)=sin^22x+sin2x\)

\(y=\left(sin2x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(\left\{{}\begin{matrix}sin^22x\le1\\sin2x\le1\end{matrix}\right.\) \(\Rightarrow y\le1+1=2\)

\(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{4}\\b=2\end{matrix}\right.\) 

\(\Rightarrow4a+b=1\)

14.

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=x+\dfrac{3\pi}{4}+k2\pi\\2x-\dfrac{\pi}{4}=\dfrac{\pi}{4}-x+k2\pi\end{matrix}\right.\)

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

\(\Rightarrow x=\left\{\dfrac{\pi}{6};\dfrac{5\pi}{6}\right\}\)

\(\Rightarrow\dfrac{\pi}{6}+\dfrac{5\pi}{6}=\pi\)

15.

\(3cosx+2cos^2x-1-cos3x+1=cosx-cos3x\)

\(\Leftrightarrow cos^2x+cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=-1\end{matrix}\right.\)

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

\(\Rightarrow\) Nghiệm lớn nhất \(x=\dfrac{3\pi}{2}\)

\(sin\left(\dfrac{3\pi}{2}-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)

16.

\(cos\left(2x+\dfrac{2\pi}{3}\right)+4cos\left(\dfrac{\pi}{6}-x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow cos\left[\pi-2\left(\dfrac{\pi}{6}-x\right)\right]+4cos\left(\dfrac{\pi}{6}-x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow-cos\left[2\left(\dfrac{\pi}{6}-x\right)\right]+4cos\left(\dfrac{\pi}{6}-x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow1-2cos^2\left(\dfrac{\pi}{6}-x\right)+4cos\left(\dfrac{\pi}{6}-x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow1-2t^2+4t=\dfrac{5}{2}\Leftrightarrow4t^2-8t+3=0\)

17.

\(sin2x=sinx\Rightarrow\left\{{}\begin{matrix}2x=x+k2\pi\\2x=\pi-x+k2\pi\end{matrix}\right.\)

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

Nghiệm dương nhỏ nhất: \(x=\dfrac{\pi}{3}\)

18.

\(-1\le sin3x\le1\Rightarrow-1\le y\le4\)

\(miny=-1\) ; \(maxy=4\)

18 ver 1

Câu này trắc nghiệm điển hình, chỉ thay \(x=40^0\) vào 4 đáp án để thử, ko ai tự luận nó cả

19.

ĐKXĐ:

\(cos\left(2x-\dfrac{\pi}{3}\right)\ne0\)

\(\Leftrightarrow2x-\dfrac{\pi}{3}\ne\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x\ne\dfrac{5\pi}{12}+\dfrac{k\pi}{2}\)

27.

\(cos\left(\dfrac{x}{2}+15^0\right)=sinx\)

\(\Leftrightarrow cos\left(\dfrac{x}{2}+15^0\right)=cos\left(90^0-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}+15^0=90^0-x+k360^0\\\dfrac{x}{2}+15^0=x-90^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=50^0+k240^0\\x=210^0+k720^0\end{matrix}\right.\)

Với \(k=1\Rightarrow x=50^0+240^0=290^0\)

28.

\(sin\left(x-\dfrac{\pi}{4}\right)\ne0\)

\(\Leftrightarrow x-\dfrac{\pi}{4}\ne k\pi\)

\(\Leftrightarrow x\ne\dfrac{\pi}{4}+k\pi\)

29.

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k2\pi\\x=-\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=\left\{\dfrac{17\pi}{12};\dfrac{23\pi}{12}\right\}\)

30.

Pt \(2sinx+3cosx=1\) có \(2^2+3^2>1^2\) nên có nghiệm

40.

\(\sqrt{3}tanx-1=0\)

\(\Leftrightarrow tanx=\dfrac{1}{\sqrt{3}}\)

\(\Rightarrow x=30^0+k180^0\)

41.

Từ đồ thị ta thấy hàm đồng biến trên \(\left(-\dfrac{\pi}{2};0\right)\) (đồ thị hàm số chỉ có xu hướng đi lên chứ không có đoạn đi xuống trên toàn miền)

42.

\(\Leftrightarrow1-cos2x-2sin2x+2\left(1+cos2x\right)=1\)

\(\Leftrightarrow cos2x-2sin2x=-2\)

43.

\(cos3x=1\Leftrightarrow4cos^3x-3cosx-1=0\)

\(\Leftrightarrow\left(cosx-1\right)\left(2cosx+1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

\(cos2x=-\dfrac{1}{2}\Leftrightarrow2cos^2x-1=-\dfrac{1}{2}\)

\(\Leftrightarrow cos^2x=\dfrac{1}{4}\Rightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

So sánh 2 pt trên ta thấy chúng có nghiệm chung \(cosx=-\dfrac{1}{2}\)

\(\Leftrightarrow x=\pm\dfrac{2\pi}{3}+k2\pi\)

11.

\(sin^2x-4sinx.cosx+3cos^2x=0\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx-3cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\sinx-3cosx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\sinx=3cosx\end{matrix}\right.\)

Với \(cosx=0\Rightarrow\) pt vô nghiệm

Với \(cosx\ne0\)

\(pt\Leftrightarrow\left[{}\begin{matrix}tanx=0\\tanx=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan3+k\pi\end{matrix}\right.\)

12.

\(pt\Leftrightarrow\sqrt{3}tanx+1=0\)

\(\Leftrightarrow tanx=-\dfrac{\sqrt{3}}{3}\)

\(\Leftrightarrow x=-\dfrac{\pi}{6}+k\pi\)