1.

\(y=\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)\Rightarrow\) tập giá trị là \(\left[-\sqrt{2};\sqrt{2}\right]\)

2. ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\)

\(\frac{cosx-sin2x}{cos2x+sinx}=\sqrt{3}\)

\(\Leftrightarrow cosx-sin2x=\sqrt{3}cos2x+\sqrt{3}sinx\)

\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{\sqrt{3}}{2}cos2x+\frac{1}{2}sin2x\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos\left(2x-\frac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=x+\frac{\pi}{3}+k2\pi\\2x-\frac{\pi}{6}=-x-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

3.

\(\Leftrightarrow2y+y.cosx=sinx+2cosx+3\)

\(\Leftrightarrow sinx+\left(2-y\right)cosx=2y-3\)

\(\Rightarrow1^2+\left(2-y\right)^2\ge\left(2y-3\right)^2\)

\(\Leftrightarrow3y^2-8y+4\le0\)

\(\Rightarrow\frac{2}{3}\le y\le2\)

4.

\(y=2\left(\frac{\sqrt{3}}{2}cos\frac{x}{2}-\frac{1}{2}sin\frac{x}{2}\right)=2cos\left(\frac{x}{2}+\frac{\pi}{6}\right)\)

\(\Rightarrow-2\le y\le2\)

5.

\(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=\frac{1}{2}sin7x-\frac{\sqrt{3}}{2}cos7x\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\left(7x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\7x-\frac{\pi}{3}=\frac{5\pi}{6}-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

6.

\(\Leftrightarrow\frac{1}{2}cos6x+\frac{1}{2}cos4x=\frac{1}{2}cos6x+\frac{1}{2}cos2x+\frac{3}{2}+\frac{3}{2}cos2x+1\)

\(\Leftrightarrow cos4x=4cos2x+5\)

\(\Leftrightarrow2cos^22x-1=4cos2x+5\)

\(\Leftrightarrow cos^22x-2cos2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3>1\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

7.

Thay lần lượt 4 đáp án ta thấy chỉ có đáp án C thỏa mãn

8.

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=\left\{\frac{\pi}{6};\frac{\pi}{2}\right\}\)

1.

\(\Leftrightarrow1-cos^22x-2\left(\frac{1+cos2x}{2}\right)+\frac{3}{4}=0\)

\(\Leftrightarrow-cos^22x-cos2x+\frac{3}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\frac{1}{2}\\cos2x=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=\pm\frac{\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\frac{\pi}{6}+k\pi\)

2.

\(2\left(2cos^2x-1\right)+2cosx-\sqrt{2}=0\)

\(\Leftrightarrow4cos^2x+2cosx-2-\sqrt{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{2}}{2}\\cosx=-\frac{1+\sqrt{2}}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k2\pi\\x=-\frac{\pi}{4}+l2\pi\end{matrix}\right.\) mà \(-\frac{\pi}{2}< x< \frac{5\pi}{2}\Rightarrow\left\{{}\begin{matrix}-\frac{\pi}{2}< \frac{\pi}{4}+k2\pi< \frac{5\pi}{2}\\-\frac{\pi}{2}< -\frac{\pi}{4}+l2\pi< \frac{5\pi}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=0;1\\l=0;1\end{matrix}\right.\) \(\Rightarrow x=\left\{\frac{\pi}{4};\frac{9\pi}{4};-\frac{\pi}{4};\frac{7\pi}{4}\right\}\)

Có 4 nghiệm

3. ĐKXĐ: ...

\(2tanx-\frac{2}{tanx}-3=0\)

\(\Leftrightarrow2tan^2x-3tanx-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-\frac{1}{2}\\tanx=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(-\frac{1}{2}\right)+k\pi\\x=arctan\left(2\right)+k\pi\end{matrix}\right.\)

Có 3 nghiệm trong khoảng đã cho \(x=arctan\left(-\frac{1}{2}\right);x=arctan\left(-\frac{1}{2}\right)+\pi;x=arctan\left(2\right)\)

4. ĐKXĐ: ...

\(\Leftrightarrow\sqrt{3}\left(1+cot^2x\right)=3cotx+\sqrt{3}\)

\(\Leftrightarrow cot^2x-\sqrt{3}cotx=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{6}+k\pi\end{matrix}\right.\)

Nghiệm âm lớn nhất của pt là \(x=-\frac{\pi}{2}\)

5. ĐKXĐ; ...

\(\Leftrightarrow tan^2x-\left(1+\sqrt{3}\right)tanx+\sqrt{3}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\sqrt{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{3}+l\pi\end{matrix}\right.\)

\(\left\{{}\begin{matrix}-2019\pi< \frac{\pi}{4}+k\pi< 2019\pi\\-2019\pi< \frac{\pi}{3}+l\pi< 2019\pi\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-2019\le k\le2018\\-2019\le l\le2018\end{matrix}\right.\)

Tổng các nghiệm: \(2.\left(-2019\pi\right)+4038\left(\frac{\pi}{3}+\frac{\pi}{4}\right)=-\frac{3365\pi}{2}< -3\)

Đáp án A đúng

1.

\(\Leftrightarrow4\left(\frac{1-cos2x}{2}\right)+3\sqrt{3}sin2x-2\left(\frac{1+cos2x}{2}\right)=4\)

\(\Leftrightarrow\sqrt{3}sin2x-cos2x=1\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=\frac{1}{2}\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=\frac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\2x-\frac{\pi}{6}=\frac{5\pi}{6}+l2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{2}+l\pi\end{matrix}\right.\)

\(\Rightarrow\) Nghiệm dương nhỏ nhất \(x=\frac{\pi}{6}\)

2.

\(\Leftrightarrow6\left(\frac{1-cos2x}{2}\right)+7\sqrt{3}sin2x-8\left(\frac{1+cos2x}{2}\right)=6\)

\(\Leftrightarrow\sqrt{3}sin2x-cos2x=1\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=\frac{1}{2}\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\2x-\frac{\pi}{6}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

3.

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

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

\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}\alpha=-\frac{\pi}{6}\\\beta=\frac{\pi}{2}\end{matrix}\right.\) \(\Rightarrow\alpha\beta=-\frac{\pi^2}{12}\)

\(cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)

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

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

\(\Rightarrow x=\left\{\frac{\pi}{2};\pi\right\}\)

a/

Đặt \(x+\frac{\pi}{3}=a\Rightarrow x=a-\frac{\pi}{3}\)

Pt trở thành:

\(cos^2a+4cos\left(\frac{\pi}{6}-a+\frac{\pi}{3}\right)=4\)

\(\Leftrightarrow cos^2a+4cos\left(\frac{\pi}{2}-a\right)-4=0\)

\(\Leftrightarrow cos^2a+4sina-4=0\)

\(\Leftrightarrow1-sin^2a+4sina-4=0\)

\(\Leftrightarrow-sin^2a+4sina-3=0\)

\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sin\left(x+\frac{\pi}{3}\right)=1\)

\(\Rightarrow x+\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=\frac{\pi}{6}+k2\pi\)

b/

Đặt \(x+\frac{\pi}{6}=a\Rightarrow x=a-\frac{\pi}{6}\)

Pt trở thành:

\(5cos2a=4sin\left(\frac{5\pi}{6}-a+\frac{\pi}{6}\right)-9\)

\(\Leftrightarrow5cos2x=4sin\left(\pi-a\right)-9\)

\(\Leftrightarrow5\left(1-2sin^2a\right)=4sina-9\)

\(\Leftrightarrow10sin^2a+4sina-14=0\)

\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=-\frac{7}{5}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=1\)

\(\Rightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=\frac{\pi}{3}+k2\pi\)

c/

\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)

\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)

\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)

\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)

\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=...\)

a/

\(\Leftrightarrow4sinx.cosx\left(sin^4x-cos^4x\right)=sin^24x\)

\(\Leftrightarrow2sin2x\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^24x\)

\(\Leftrightarrow-2sin2x.cos2x=sin^24x\)

\(\Leftrightarrow-sin4x=sin^24x\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin4x=-1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=-\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)

b/

\(\Leftrightarrow2\left(1-cosx\right)-\sqrt{3}cos2x=1+1+cos\left(2x-\frac{3\pi}{2}\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=sin\left(2\pi-2x\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)

\(\Leftrightarrow sin2x-\sqrt{3}cos2x=2cosx\)

\(\Leftrightarrow\frac{1}{2}sin2x-\sqrt{3}cos2x=cosx\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=cosx=sin\left(\frac{\pi}{2}-x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{5}{4}=0\)

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{3}\right)-\frac{5}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\\sin\left(x+\frac{\pi}{3}\right)=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

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

\(\Leftrightarrow cosx-2cosx+m=0\)

\(\Leftrightarrow cosx=-m\)

Từ đường tròn lượng giác ta thấy để pt có đúng 3 nghiệm pb thuộc \(\left[0;\frac{7\pi}{2}\right]\Rightarrow0\le-m< 1\Rightarrow-1< m\le0\)