(C): (x - 2)² + (y - 1)² = 16

=> R=4 là bán kính (C) và I(2,1) là tâm của (C)

T\(\overrightarrow{v}\) (C) --->(C') => R'=4 là tâm (C') và T\(\overrightarrow{v}\) I(2,1) --->I'(2+1,1+2)=(3,3) là tâm của (C')

Vậy (C'): (x-3)²+(y-3)²=16

bài 1:

a)\(\sin x=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.k\in Z}\)

b)\(\cos x=\dfrac{\sqrt{2}}{2}\Leftrightarrow x=\pm\dfrac{\pi}{4}+k2\pi\left(k\in Z\right)\)

Bài 2:

a)\(\sqrt{3}\cos3x-\sin3x=-1\)

\(\Leftrightarrow\dfrac{1}{2}\left(\sqrt{3}\cos3x-\sin3x\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}\cos3x-\dfrac{1}{2}\sin3x=\dfrac{-1}{2}\)

\(\Leftrightarrow\cos\dfrac{\pi}{6}\cos3x-\sin\dfrac{\pi}{6}\sin3x=\dfrac{-1}{2}\)

\(\Leftrightarrow\cos\left(\dfrac{\pi}{6}+3x\right)=\dfrac{-1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.k\in Z}\)

b)\(2\sin x+\cos x=1\)

\(\Leftrightarrow\dfrac{2}{\sqrt{2^2+1^2}}\cos x+\dfrac{1}{\sqrt{2^2+1^2}}\sin x=\dfrac{1}{\sqrt{2^2+1^2}}\)

\(\Leftrightarrow\dfrac{2}{\sqrt{5}}\cos x+\dfrac{1}{\sqrt{5}}\sin x=\dfrac{1}{\sqrt{5}}\)(*)

Đặt \(\dfrac{2}{\sqrt{5}}=\cos\alpha,\dfrac{1}{\sqrt{5}}=\sin\alpha\)

Từ phương trình (*) ta có:

\(\cos\alpha\cos x+\sin\alpha\sin x=\sin\alpha\)

\(\Leftrightarrow\cos\left(\alpha-x\right)=\sin\alpha\)

\(\Leftrightarrow\cos\left(\alpha-x\right)=\cos\left(\dfrac{\pi}{2}-\alpha\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{2}+2\alpha+k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.k\in Z}\)

Bài 3:

(C) \(\left(x+2\right)^2+\left(y-3\right)^2=16\)

\(\Leftrightarrow\left\{{}\begin{matrix}R=4\\I\left(-2,3\right)\end{matrix}\right.\)

\(V_{\left(A,\dfrac{3}{2}\right)}C\rightarrow C'\Rightarrow\left\{{}\begin{matrix}R'=\dfrac{3}{2}R=6\\V_{\left(A,\dfrac{3}{2}\right)}I\rightarrow I'\circledast\end{matrix}\right.\)

Từ \(\circledast\Rightarrow I'\left(-\dfrac{7}{2},\dfrac{7}{2}\right)\)là tâm của đường tròn C'

Vậy phương trình đường tròn (C') cần tìm là:

\(\left(x+\dfrac{7}{2}\right)^2+\left(y-\dfrac{7}{2}\right)=36\)