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

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

Với \(x\in\left(-\frac{\pi}{4};\frac{\pi}{2}\right)\Rightarrow cosx>0\Rightarrow3cosx+1>0\)

Do đó pt tương đương:

\(2cos2x-1=0\Rightarrow cos2x=\frac{1}{2}\)

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

Pt có 2 nghiệm thuộc khoảng đã cho là \(x=\left\{-\frac{\pi}{6};\frac{\pi}{6}\right\}\)

Câu 3 sai đề hả bạn

1/ ĐKXĐ: \(\cos2x\ne0\)

\(\frac{\cos4x}{\cos2x}=\frac{\sin2x}{\cos2x}\)\(\Leftrightarrow\cos4x-\sin2x=0\)

\(\Leftrightarrow2\cos^22x-1-\sin2x=0\)

\(\Leftrightarrow2-2\sin^22x-1-\sin2x=0\)

\(\Leftrightarrow2\sin^22x+\sin2x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\frac{1}{2}=\sin\frac{\pi}{6}\\\sin2x=-1=\sin\frac{-\pi}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)

2/ \(\sin2.4x+\cos4x=1+2\sin2x.\cos\left(2x+4x\right)\)

\(\Leftrightarrow2\sin4x.\cos4x+\cos4x=1+2\sin2x.\left(\cos2x.\cos4x-\sin2x.\sin4x\right)\)

\(\Leftrightarrow2\sin4x.\cos4x+\cos4x=1+2\sin2x.\cos2x.\cos4x-2\sin^22x.\sin4x\)

\(\Leftrightarrow2\sin4x.\cos4x+\cos4x=1+\sin4x.\cos4x-\sin4x+\cos4x.\sin4x\)

Đến đây bn tự giải nốt nhé, lm kiểu bthg thôi bởi vì đã quy về hết sin4x và cos4x r