a, \(sin5x-cos5x+1=0\)
\(\Leftrightarrow\sqrt{2}sin\left(5x-\dfrac{\pi}{4}\right)+1=0\)
\(\Leftrightarrow sin\left(5x-\dfrac{\pi}{4}\right)=-\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\5x-\dfrac{\pi}{4}=\pi+\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{5}\\x=\dfrac{3\pi}{10}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
b, \(sin3x+cos3x=1\)
\(\Leftrightarrow\sqrt{2}sin\left(3x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(3x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\3x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
c, \(2sin5x=\sqrt{6}-2cos5x\)
\(\Leftrightarrow2sin5x+2cos5x=\sqrt{6}\)
\(\Leftrightarrow2\sqrt{2}cos\left(5x-\dfrac{\pi}{4}\right)=\sqrt{6}\)
\(\Leftrightarrow cos\left(5x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow5x-\dfrac{\pi}{4}=\pm\dfrac{\pi}{6}+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{60}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
d, \(3cos2x-4sin2x=1\)
\(\Leftrightarrow\dfrac{3}{5}cos2x-\dfrac{4}{5}sin2x=\dfrac{1}{5}\)
\(\Leftrightarrow cos\left(2x+arccos\dfrac{3}{5}\right)=\dfrac{1}{5}\)
\(\Leftrightarrow2x+arccos\left(\dfrac{3}{5}\right)=\pm arccos\left(\dfrac{1}{5}\right)+k2\pi\)
\(\Leftrightarrow x=-\dfrac{1}{2}arccos\left(\dfrac{3}{5}\right)\pm\dfrac{1}{2}arccos\left(\dfrac{1}{5}\right)+k\pi\)
e, \(sinx-\sqrt{3}cosx=1\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)
f, \(\sqrt{3}sin2x+cos2x=2\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x+\dfrac{1}{2}cos2x=1\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=1\)
\(\Leftrightarrow2x+\dfrac{\pi}{3}=k2\pi\)
\(\Leftrightarrow x=-\dfrac{\pi}{6}+k\pi\)
