(C) có \(\left\{{}\begin{matrix}I\left(4,0\right)\\R=4\end{matrix}\right.\)

\(T_{\overrightarrow{v}}\left(I\right)=I'(x',y')\Rightarrow\left\{{}\begin{matrix}x'=x+a=4+3=7\\y'=y+a=0+\left(-1\right)=-1\end{matrix}\right.\)   \(\Rightarrow I'\left(7,-1\right)\)

\(T_{\overrightarrow{v}}\left(C\right)=\left(C'\right)\) có tâm \(I'\left(7,-1\right)R=4\)

  \(\Rightarrow\left(C'\right):\left(x-7\right)^2+\left(y+1\right)^2=4\)

(C) có \(\left\{{}\begin{matrix}I\left(2;3\right)\\R=3\end{matrix}\right.\)

\(T_{\overrightarrow{v}}\left(I\right)=I'\left(x',y'\right)\)\(\Rightarrow\left\{{}\begin{matrix}x'=x+a=2+\left(-1\right)=1\\y'=y+a=3+2=5\end{matrix}\right.\Rightarrow I'\left(1,5\right)\)

\(T_{\overrightarrow{v}}\left(C\right)=\left(C'\right)\) có \(\left\{{}\begin{matrix}I'\left(1,5\right)\\R=3\end{matrix}\right.\)

\(\Rightarrow\) \(\left(C'\right):\left(x-1\right)^2+\left(y-5\right)^2=3\)

A(x_A,y_A)

\(Q_{\left(O,\dfrac{\pi}{2}\right)}\left(A\right)=C\)       \(\Rightarrow C\left(-y_A,x_A\right)\)

a, \(sin\left(2x+45^o\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+45^o=-30^o+k.360^o\\2x+45^o=-150^o+k.360^o\end{matrix}\right.\)

...

b, \(cos\left(4x+\dfrac{\pi}{5}\right)-sin2x=0\)

\(\Leftrightarrow cos\left(4x+\dfrac{\pi}{5}\right)-cos\left(\dfrac{\pi}{2}-2x\right)=0\)

\(\Leftrightarrow-2sin\left(x+\dfrac{\pi}{20}\right).sin\left(3x-\dfrac{3\pi}{20}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{20}\right)=0\\sin\left(3x-\dfrac{3\pi}{20}\right)=0\end{matrix}\right.\)

...

a) \(sinx=\dfrac{1}{2}\Rightarrow sinx=sin\dfrac{\pi}{6}\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

b) Pt: \(\Rightarrow cos\left(x+\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{3}}{2}=cos\dfrac{5\pi}{6}\)

....bạn tự tìm x nhé!

c)Pt: \(\Rightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-4\end{matrix}\right.\)  (loại sinx=-4 vì sinx\(\in[-1,1]\)

Cảm ơn

(m+2)sinx+mcosx=2(m+2)sin⁡x+mcos⁡x=2

Phương trình có nghiệm

⇔(m+2)2+m2≥22⇔2m2+4m+4≥4⇔m2+2m≥0⇔[m≥0m≤−2⇔(m+2)2+m2≥22⇔2m2+4m+4≥4⇔m2+2m≥0⇔[m≥0m≤−2

b) Phương trình (2m−1)sinx+(m−1)cosx=m−3(2m−1)sinx+(m−1)cosx=m−3 vô nghiệm

⇔(2m−1)2+(m−1)2≤(m−3)2⇔4m2−4m+1+m2−2m+1≤m2−6m+9⇔4m2≤7⇔m2≤74⇔−√72≤m≤√72