Hàm là vậy phải không nhỉ? \(y=\dfrac{sin^2x-3sinx}{\left(tanx-1\right)\left(cotx+1\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\tanx-1\ne0\\cotx+1\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\tanx\ne1\\cotx\ne-1\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{4}+k\pi\\x\ne-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ne\dfrac{k\pi}{4}\)

\(sin^2x+sin^22x=1\)

\(\Leftrightarrow2sin^2x-1+2sin^22x-2=-1\)

\(\Leftrightarrow-cos2x-2cos^22x+1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)

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

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

ĐKXĐ: ...

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

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

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

\(\Leftrightarrow...\)

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) 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)

1/ \(sinx=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\)

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

b/ \(cos=-\frac{\sqrt{2}}{2}=cos\left(\frac{3\pi}{4}\right)\)

\(\Rightarrow x=\pm\frac{3\pi}{4}+k2\pi\)

c/ \(tanx=\sqrt{3}=tan\left(\frac{\pi}{3}\right)\)

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

d/ \(cotx=0\Rightarrow x=\frac{\pi}{2}+k\pi\)

2/

a/ \(sin^2x+sinx-2=0\)

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

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

b/ \(cot^2x-2cotx-3=0\)

\(\Leftrightarrow\left(cotx+1\right)\left(cotx-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cotx=-1\\cotx=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=arccot3+k\pi\end{matrix}\right.\)

3/ \(\Leftrightarrow1-cos2x+1-cos4x+1-cos6x=3\)

\(\Leftrightarrow cos2x+cos6x+cos4x=0\)

\(\Leftrightarrow2coss4x.cos2x+cos4x=0\)

\(\Leftrightarrow cos4x\left(2cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\frac{2\pi}{3}+k2\pi\\2x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)

a ) \(2cosx-3sinx+2=0\) 

\(\Leftrightarrow2cosx-3sinx=-2\)  

\(\Leftrightarrow\dfrac{2}{\sqrt{13}}cosx-\dfrac{3}{\sqrt{13}}sinx=-\dfrac{2}{\sqrt{13}}\) 

Thấy : \(\left(\dfrac{2}{\sqrt{13}}\right)^2+\left(\dfrac{-3}{\sqrt{13}}\right)^2=1\) nên tồn tại \(\alpha\) t/m : 

\(sin\alpha=\dfrac{2}{\sqrt{13}};cos\alpha=\dfrac{-3}{\sqrt{13}}\) . . Khi đó : \(sin\alpha.cosx+cos\alpha.sinx=\dfrac{-2}{\sqrt{13}}\)

\(\Leftrightarrow sin\left(\alpha+x\right)=\dfrac{-2}{\sqrt{13}}\) ( p/t cơ bản ) 

b ) \(\dfrac{1+sinx}{1+cosx}=\dfrac{1}{2}\) ( ĐK : \(cosx\ne-1\Leftrightarrow x\ne\left(2k+1\right)\pi\) ; ( k thuộc Z )  ) 

\(\Leftrightarrow2+2sinx=cosx+1\) \(\Leftrightarrow cosx-2sinx=1\) 

Làm giống như a )  

c ) \(cos\left(2x-15^o\right)+sin\left(2x-15^o\right)=-1\)  

Đặt \(t=2x-15^o\) ; ta có : \(cos t + sin t = -1\) 

\(\Leftrightarrow\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)=-1\) \(\Leftrightarrow sin\left(t+\dfrac{\pi}{4}\right)=sin\left(-\dfrac{\pi}{4}\right)\)  

Xong rồi bn làm tiếp ; chú ý đổi ra độ