Question

Giải pt lượng giác Giúp em với em cảm ơn ạ

Answer 1

1.

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

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

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

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

Answer 2

4.

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

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=sin\left(-\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)

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

Answer 3

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

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

     \(\Rightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{2\pi}{3}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)  \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{6}+k2\pi\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)

2) \(2sin\left(x+\dfrac{\pi}{4}\right)+\sqrt{2}=0\)

\(\Rightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow sin\left(x+\dfrac{\pi}{4}\right)=sin\left(-\dfrac{\pi}{4}\right)\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi+\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)