Question

b. Cos3x - sin3x = - căn2
c. Căn3 sin2x + cos 2x =-căn 2

Answer 1

b.

\(\Leftrightarrow\sqrt{2}cos\left(3x+\frac{\pi}{4}\right)=-\sqrt{2}\)

\(\Leftrightarrow cos\left(3x+\frac{\pi}{4}\right)=-1\)

\(\Leftrightarrow3x+\frac{\pi}{4}=\pi+k2\pi\)

\(\Leftrightarrow x=...\)

c.

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=-\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=-\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=-\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{6}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)