Question

sin(4x+pi/4)=sin(x-pi/5)

Answer 1

TA có: \(\sin\left(4x+\frac{\pi}{4}\right)=\sin\left(x-\frac{\pi}{5}\right)\)

=>\(\left[\begin{array}{l}4x+\frac{\pi}{4}=x-\frac{\pi}{5}+k2\pi\\ 4x+\frac{\pi}{4}=\pi-x+\frac{\pi}{5}+k2\pi=-x+\frac65\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}4x-x=-\frac{\pi}{5}-\frac{\pi}{4}+k2\pi\\ 4x+x=\frac65\pi-\frac{\pi}{4}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=-\frac{9}{20}\pi+k2\pi\\ 5x=\frac{19}{20}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{3}{20}\pi+\frac{k2\pi}{3}\\ x=\frac{19}{100}\pi+\frac{k2\pi}{5}\end{array}\right.\)