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.

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

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

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

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

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

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

\(\Rightarrow\) 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.

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

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

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

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

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

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

1.

Đề là \(x\in\left(0;\frac{\pi}{4}\right)\) hay \(x\in\left[0;\frac{\pi}{4}\right]\) ?

2.

\(sin3x-4sinx.cos2x=0\)

\(\Leftrightarrow sin3x-\left(2sin3x-2sinx\right)=0\)

\(\Leftrightarrow2sinx-sin3x=0\)

\(\Leftrightarrow2sinx-3sinx+4sin^3x=0\)

\(\Leftrightarrow sinx\left(4sin^2x-1\right)=0\)

\(\Leftrightarrow sinx\left(1-2cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos2x=\frac{1}{2}\end{matrix}\right.\)

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

3.

\(sin^2x.cosx=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

4.

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

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

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

\(\Leftrightarrow2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{\pi}{3}+k\pi\)

5.

Ko có 4 đáp án thì làm sao biết, có vô số pt tương đương với pt này :)

6.

\(sinx+cosx-2sinx.cosx+1=0\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(t+1-t^2+1=0\)

\(\Leftrightarrow-t^2+t+2=0\)

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

\(\Rightarrow2sinx.cosx=t^2-1=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

\(\sqrt{3}sin2x-\left(1+cos2x\right)=m\)

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

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

Do \(x\in\left[\frac{\pi}{4};\frac{5\pi}{12}\right]\Rightarrow2x-\frac{\pi}{6}\in\left[\frac{\pi}{3};\frac{2\pi}{3}\right]\)

\(\Rightarrow sin\left(2x-\frac{\pi}{6}\right)\in\left[\frac{\sqrt{3}}{2};1\right]\)

Pt có nghiệm thuộc đoạn đã cho khi và chỉ khi: \(\frac{\sqrt{3}}{2}\le\frac{m+1}{2}\le1\)

\(\Leftrightarrow\sqrt{3}-1\le m\le1\)

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

\(\Leftrightarrow cosx=-m\)

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

1.

\(\Leftrightarrow2cos2x+\sqrt{2}.\frac{\sqrt{2}}{2}=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)

\(\Rightarrow x=\left\{\frac{\pi}{3};\frac{4\pi}{3};\frac{2\pi}{3};\frac{5\pi}{3}\right\}\)

2.

\(\Leftrightarrow sin4x-cos4x+sin4x+cos4x=\sqrt{6}\)

\(\Leftrightarrow2sin4x=\sqrt{6}\)

\(\Leftrightarrow sin4x=\frac{\sqrt{6}}{2}>1\)

Pt vô nghiệm

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/

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

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

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

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

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