Question

[Thùy Dương]

Mn giải giúp em với ạ,e cảm ơn nhiều

Answer 1

[Hồng Phúc]

1.

c, \(sin\left(\dfrac{\pi}{3}-x\right)=-\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{3}-x=arcsin\left(-\dfrac{1}{4}\right)+k.360^o\\\dfrac{\pi}{3}-x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k.360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}-arcsin\left(-\dfrac{1}{4}\right)+k.360^o\\x=-\dfrac{2\pi}{3}+arcsin\left(-\dfrac{1}{4}\right)+k.360^o\end{matrix}\right.\)

d, \(sin4x=\dfrac{2}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=arcsin\dfrac{2}{3}+k2\pi\\4x=\pi-arcsin\dfrac{2}{3}+k2\pi\end{matrix}\right.\)

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

Answer 2

[Hồng Phúc]

1.

e, \(2sin2x+\sqrt{2}=0\)

\(\Leftrightarrow sin2x=-\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin2x=sin\left(-\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{4}+k2\pi\\2x=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+k\pi\\x=\dfrac{5\pi}{8}+k\pi\end{matrix}\right.\)