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.

\(\Leftrightarrow1-2sin^2x+sinx+m=0\)

\(\Leftrightarrow2sin^2x-sinx-1=m\)

Đặt \(sinx=t\Rightarrow t\in\left[-\dfrac{1}{2};\dfrac{\sqrt{2}}{2}\right]\)

Xét hàm \(f\left(t\right)=2t^2-t-1\) trên \(\left[-\dfrac{1}{2};\dfrac{\sqrt{2}}{2}\right]\)

\(-\dfrac{b}{2a}=\dfrac{1}{4}\in\left[-\dfrac{1}{2};\dfrac{\sqrt{2}}{2}\right]\)

\(f\left(-\dfrac{1}{2}\right)=0\) ; \(f\left(\dfrac{1}{4}\right)=-\dfrac{9}{8}\) ; \(f\left(\dfrac{\sqrt{2}}{2}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow-\dfrac{9}{8}\le f\left(t\right)\le0\Rightarrow-\dfrac{9}{8}\le m\le0\)

Có 2 giá trị nguyên của m (nếu đáp án là 3 thì đáp án sai)

2.

ĐKXĐ: \(sin2x\ne1\Rightarrow x\ne\dfrac{\pi}{4}\) (chỉ quan tâm trong khoảng xét)

Pt tương đương:

\(\left(tan^2x+cot^2x+2\right)-\left(tanx+cotx\right)-4=0\)

\(\Leftrightarrow\left(tanx+cotx\right)^2+\left(tanx+cotx\right)-4=0\)

\(\Rightarrow\left[{}\begin{matrix}tanx+cotx=\dfrac{1+\sqrt{17}}{2}\\tanx+cotx=\dfrac{1-\sqrt{17}}{2}\left(loại\right)\end{matrix}\right.\)

Nghiệm xấu quá, kiểm tra lại đề chỗ \(-tanx+...-cotx\) có thể 1 trong 2 cái đằng trước phải là dấu "+"

Miền \(\left[-\dfrac{\pi}{3};\dfrac{\pi}{2}\right]\) là cung tròn CAB

Chiếu cung tròn lên trục cos (trục ngang) được đoạn màu đỏ, với O có hoành độ bằng 0, A có hoành độ bằng 1

Do đó miền giá trị của cos trên \(\left[-\dfrac{\pi}{3};\dfrac{\pi}{2}\right]\) là \(\left[0;1\right]\) hay đoạn OA

\(\Leftrightarrow2x+\frac{\pi}{3}=\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{2}\)

Do \(x\in\left[0;2\pi\right]\Rightarrow0\le\frac{\pi}{12}+\frac{k\pi}{2}\le2\pi\)

\(\Rightarrow-\frac{1}{6}\le k\le\frac{23}{6}\Rightarrow k=\left\{0;1;2;3\right\}\)

\(\Rightarrow x=\left\{\frac{\pi}{12};\frac{7\pi}{12};\frac{13\pi}{12};\frac{19\pi}{12}\right\}\)

a) Pt\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2xcos^2x\left(sin^2x+cos^2x\right)+3sinx.cosx-\dfrac{m}{4}+2=0\)

\(\Leftrightarrow1-\dfrac{3}{4}sin^22x-\dfrac{3}{2}sin2x-\dfrac{m}{4}+2=0\)

\(\Leftrightarrow-3sin^22x-6sin2x-m+12=0\)

Đặt \(t=sin2x;t\in\left[-1;1\right]\)

Pttt: \(-3t^2-6t-m+12=0\)

\(\Leftrightarrow-3t^2-6t+12=m\) (1)

Đặt \(f\left(t\right)=-3t^2-6t+12;t\in\left[-1;1\right]\) 

Vẽ BBT sẽ tìm được \(f\left(t\right)_{min}=3;f\left(t\right)_{max}=15\)\(\Leftrightarrow3\le f\left(t\right)\le15\)\(\Rightarrow m\in\left[3;15\right]\) thì pt (1) sẽ có nghiệm

mà \(m\in Z\) nên tổng m nguyên để pt có nghiệm là 13 m

Vậy có tổng 13 m nguyên

b) Pt\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\left(1\right)\\2cos^2x-\left(2m+1\right)cosx+m=0\left(2\right)\end{matrix}\right.\)

Từ (1)\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)

\(x\in\left[0;2\pi\right]\Rightarrow0\le\dfrac{\pi}{2}+k2\pi\le2\pi\)\(\Leftrightarrow-\dfrac{1}{4}\le k\le\dfrac{3}{4}\)\(\Rightarrow k=0\)

Tại k=0\(\Rightarrow x=\dfrac{\pi}{2}\)

Để pt ban đầu có 4 nghiệm pb \(\in\left[0;2\pi\right]\)

\(\Leftrightarrow\) Pt (2) có 3 nghiệm pb khác \(\dfrac{\pi}{2}\)

Xét pt (2) có: \(2cos^2x-\left(2m+1\right)cosx+m=0\)

Vì là phương trình bậc hai ẩn \(cosx\) nên pt (2) chỉ có nhiều nhất ba nghiệm \(\Leftrightarrow\) Pt (2) có một nghiệm cosx=0

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) mà \(x\ne\dfrac{\pi}{2}\)

\(\Rightarrow\) Pt (2) chỉ có nhiều nhất hai nghiệm

\(\Rightarrow\) Pt ban đầu không thể có 4 nghiệm phân biệt

Vậy \(m\in\varnothing\) 

\(\sqrt{2}cos\left(x+\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=\pm\dfrac{\pi}{4}+k2\pi\)

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

\(0\le-\dfrac{\pi}{12}+k2\pi\le2\pi\Leftrightarrow...\)

\(0\le-\dfrac{7\pi}{12}+k2\pi\le2\pi\Leftrightarrow...\)

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.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=-\frac{\pi}{3}+k2\pi\\x+\frac{\pi}{6}=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)

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

Do \(x\in\left[0;2\pi\right]\Rightarrow\left[{}\begin{matrix}0\le-\frac{\pi}{2}+k2\pi\le2\pi\\0\le\frac{7\pi}{6}+k2\pi\le2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}k=1\\k=0\end{matrix}\right.\) \(\Rightarrow x=\left\{\frac{3\pi}{2};\frac{7\pi}{6}\right\}\)

\(sin3x=0\Leftrightarrow3x=k\pi\)

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

Do \(x\in\left[0;2\pi\right]\Rightarrow0\le\frac{k\pi}{3}\le2\pi\Rightarrow0\le k\le6\)

\(\Rightarrow x=\left\{0;\frac{\pi}{3};\frac{2\pi}{3};\pi;\frac{4\pi}{3};\frac{5\pi}{3};2\pi\right\}\)

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