a/

ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow3tanx-\sqrt{3}=0\)

\(\Rightarrow tanx=\frac{1}{\sqrt{3}}\)

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

b/

ĐKXĐ: \(sinx\ne-1\)

\(\Leftrightarrow\frac{\left(2-sinx\right)\left(\sqrt{3}cosx-1\right)}{1+sinx}+2-sinx=0\)

\(\Leftrightarrow\left(2-sinx\right)\left(\frac{\sqrt{3}cosx-1}{1+sinx}+1\right)=0\)

\(\Leftrightarrow\frac{\sqrt{3}cosx-1}{1+sinx}=-1\) (do 2-sinx>0 với mọi x)

\(\Leftrightarrow\sqrt{3}cosx-1=-1-sinx\)

\(\Leftrightarrow sinx=-\sqrt{3}cosx\Rightarrow tanx=-\sqrt{3}\)

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

c/

ĐKXĐ: \(sin2x\ne0\)

\(\Leftrightarrow\frac{\frac{sinx}{cosx}-sinx}{sin^3x}=\frac{1}{cosx}\)

\(\Leftrightarrow sinx-sinx.cosx=sin^3x\)

\(\Leftrightarrow1-cosx=sin^2x\)

\(\Leftrightarrow1-cosx=1-cos^2x\)

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

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

d/

ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow\frac{sin\left(3x-x\right)}{cos^2x}=2\sqrt{3}\)

\(\Leftrightarrow\frac{sin2x}{cos^2x}=2\sqrt{3}\)

\(\Leftrightarrow\frac{2sinx.cosx}{cos^2x}=2\sqrt{3}\)

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

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

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

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

a/

\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)

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

b/ ĐKXĐ: \(x\ne\frac{k\pi}{2}\)

\(\Leftrightarrow\frac{sin2x.sinx+cos2x.cosx}{sinx.cosx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}\)

\(\Leftrightarrow\frac{cos\left(2x-x\right)}{sinx.cosx}=\frac{sin^2x-cos^2x}{sinx.cosx}\)

\(\Leftrightarrow cosx=sin^2x-cos^2x\)

\(\Leftrightarrow cosx=1-2cos^2x\)

\(\Leftrightarrow2cos^2x+cosx-1=0\)

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

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

c/ ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow\frac{1}{cos^2x}=\frac{1-cos^2x+1-sin^3x}{1-sin^3x}\)

\(\Leftrightarrow\frac{1}{cos^2x}=\frac{sin^2x}{1-sin^3x}+1\)

\(\Leftrightarrow\frac{1}{cos^2x}-1=\frac{sin^2x}{1-sin^3x}\)

\(\Leftrightarrow\frac{1-cos^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)

\(\Leftrightarrow\frac{sin^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\cos^2x=1-sin^3x\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow1-sin^2x=1-sin^3x\)

\(\Leftrightarrow sin^3x-sin^2x=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=1\left(l\right)\end{matrix}\right.\)

a.

ĐKXĐ: \(sin2x\ne0\Leftrightarrow x\ne\frac{k\pi}{2}\)

\(sinx+cosx=\frac{2cosx}{sinx}-\frac{2sinx}{cosx}\)

\(\Leftrightarrow sinx+cosx=\frac{2\left(cos^2x-sin^2x\right)}{sinx.cosx}\)

\(\Leftrightarrow sinx+cosx=\frac{2\left(sinx+cosx\right)\left(cosx-sinx\right)}{sinx.cosx}\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow...\\\frac{2\left(cosx-sinx\right)}{sinx.cosx}=1\left(1\right)\end{matrix}\right.\)

Xét (1) \(\Leftrightarrow2\left(cosx-sinx\right)=sinx.cosx\)

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

\(\Rightarrow2t=\frac{1-t^2}{2}\Leftrightarrow t^2-4t-1=0\)

\(\Rightarrow\left[{}\begin{matrix}t=2+\sqrt{5}\left(l\right)\\t=2-\sqrt{5}\end{matrix}\right.\)

\(\Rightarrow cosx-sinx=2-\sqrt{5}\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{5}-2}{\sqrt{2}}=sina\)

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

a/

\(\Leftrightarrow tanx=-tan\left(\frac{2\pi}{3}-3x\right)\)

\(\Leftrightarrow tanx=tan\left(3x-\frac{2\pi}{3}\right)\)

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

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

b/

\(tan\left(2x-15^0\right)=tanx\)

\(\Rightarrow2x-15^0=x+k180^0\)

\(\Rightarrow x=15^0+k180^0\)

c/

ĐKXĐ: ...

\(\Leftrightarrow tan2x-2=3\left(2tan2x+1\right)\)

\(\Leftrightarrow5tan2x=-5\)

\(\Rightarrow tan2x=-1\)

\(\Rightarrow2x=-\frac{\pi}{4}+k\pi\)

\(\Rightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)

d/

ĐKXĐ: ...

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

\(\Leftrightarrow2sinx=2\sqrt{3}cosx\)

\(\Rightarrow tanx=\sqrt{3}\Rightarrow x=\frac{\pi}{3}+k\pi\)