Question

Answer 1

a.

\(\Leftrightarrow\sqrt{2}sin\left(5x-\dfrac{\pi}{4}\right)=-1\)

\(\Leftrightarrow sin\left(5x-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{5}\\x=\dfrac{3\pi}{10}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

Answer 2

b.

\(\Leftrightarrow\sqrt{2}sin\left(3x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(3x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

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

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

Answer 3

c.

\(\Leftrightarrow2\left(sin5x+cos5x\right)=\sqrt{6}\)

\(\Leftrightarrow2\sqrt{2}sin\left(5x+\dfrac{\pi}{4}\right)=\sqrt{6}\)

\(\Leftrightarrow sin\left(5x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+\dfrac{\pi}{4}=\dfrac{\pi}{3}+k2\pi\\5x+\dfrac{\pi}{4}=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{60}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{12}+\dfrac{k2\pi}{5}\end{matrix}\right.\)