a/

ĐKXĐ: ..

\(\Leftrightarrow1+cot^2x=cotx+3\)

\(\Leftrightarrow cot^2x-cotx-2=0\)

\(\Rightarrow\left[{}\begin{matrix}cotx=-1\\cotx=2\end{matrix}\right.\)

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

b/

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

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

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

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

c/

ĐKXĐ: ...

\(\Leftrightarrow9-13cosx+4.cos^2x=0\)

\(\Leftrightarrow\left(cosx-1\right)\left(4cosx-9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{9}{4}>1\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=k2\pi\)

d/

\(\Leftrightarrow2\left(tan^2x+1\right)+1=\frac{3}{cosx}\)

\(\Leftrightarrow\frac{2}{cos^2x}-\frac{3}{cosx}+1=0\)

\(\Leftrightarrow\left(\frac{1}{cosx}-1\right)\left(\frac{2}{cosx}-1\right)=0\)

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

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

\(\Rightarrow x=k2\pi\)

a/

\(\Leftrightarrow2cos2x.cosx+\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right).cos2x=0\)

\(\Leftrightarrow2cos2x.cosx+cos^22x=0\)

\(\Leftrightarrow cos2x\left(2cosx+cos2x\right)=0\)

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

\(\left(1\right)\Leftrightarrow2x=\frac{\pi}{2}+k\pi\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(\left(2\right)\Leftrightarrow2cosx+2cos^2x-1=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}-1}{2}\\cosx=\frac{-\sqrt{3}-1}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm arccos\left(\frac{\sqrt{3}-1}{2}\right)+k2\pi\)

b/

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

\(\Leftrightarrow cos^2x+\frac{1}{2}cosx=0\)

\(\Leftrightarrow cosx\left(cosx+\frac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

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

c/ ĐKXĐ: ...

\(\Leftrightarrow\left(\frac{sinx}{cosx}+\frac{cosx}{sinx}\right)^2+\frac{3}{sin2x}-7=0\)

\(\Leftrightarrow\left(\frac{sin^2x+cos^2x}{sinx.cosx}\right)^2+\frac{3}{sin2x}-7=0\)

\(\Leftrightarrow\left(\frac{2}{sin2x}\right)^2+\frac{3}{sin2x}-7=0\)

Đặt \(\frac{1}{sin2x}=a\Rightarrow4a^2+3a-7=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{7}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{1}{sin2x}=1\\\frac{1}{sin2x}=-\frac{7}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{4}{7}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\\2x=arcsin\left(-\frac{4}{7}\right)+k2\pi\\2x=\pi-arcsin\left(-\frac{4}{7}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(-\frac{4}{7}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(-\frac{4}{7}\right)+k\pi\end{matrix}\right.\)

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) TH1: sinx = 1 

--> x = pi/2 + k2pi (k nguyên)

TH2: sinx = -3 (loại)

b) 2cosx + cos2x = 0

<=> 2cosx + 2cos^2(x) - 1 = 0

TH1: cosx = (-1 + sqrt(3))/2

TH2: cosx = (-1 - sqrt(3))/2 (loại)

c) ĐKXĐ: x # kpi

Pt <=> tanx + 1/tanx + 2 = 0

--> tanx = -1

--> x = -pi/4 + kpi (k nguyên)

a/

\(\Leftrightarrow4sin^3x+6\sqrt{2}sinx.cosx-8sinx=0\)

\(\Leftrightarrow2sinx\left(2sin^2x+3\sqrt{2}cosx-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\2sin^2x+3\sqrt{2}cosx-4=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\left(1-cos^2x\right)+3\sqrt{2}cosx-4=0\)

\(\Leftrightarrow-2cos^2x+3\sqrt{2}cosx-2=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=\sqrt{2}>1\left(l\right)\\cosx=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow x=\pm\frac{\pi}{4}+k2\pi\)

b/

\(\Leftrightarrow4cos^3x+8sinx.cosx-7cosx=0\)

\(\Leftrightarrow cosx\left(4cos^2x+8sinx-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\4cos^2x+8sinx-7=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow4\left(1-sin^2x\right)+8sinx-7=0\)

\(\Leftrightarrow-4sin^2x+8sinx-3=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=\frac{3}{2}\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)

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

c/

ĐKXĐ; ...

\(\Leftrightarrow\frac{sinx}{cosx}+\frac{cosx}{sinx}-5+\frac{3}{sin^22x}=0\)

\(\Leftrightarrow\frac{sin^2x+cos^2x}{sinx.cosx}-5+\frac{3}{sin^22x}=0\)

\(\Leftrightarrow\frac{3}{sin^22x}+\frac{2}{sin2x}-5=0\)

Đặt \(\frac{1}{sin2x}=t\Rightarrow3t^2+2t-5=0\)

\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{5}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{1}{sin2x}=1\\\frac{1}{sin2x}=-\frac{5}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{3}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(-\frac{3}{5}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(-\frac{3}{5}\right)+k\pi\end{matrix}\right.\)