(√3-1)sinx-(√3+1)cosx+√3-1=0
⇔sin3x-√3cos3x=1
⇔$\frac{1}{3}$ sin3x+$\frac{√3}{2}$cos3x=$\frac{1}{2}$
⇔sin(3x+$\frac{π}{3}$)=$\frac{1}{2}$
⇔\(\left[ \begin{array}{l}3x+ \frac{π}{3}= \frac{π}{6} +k2π\\3x+ \frac{π}{3} = \frac{5π}{6}+k2π\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-\frac{π}{18} +k\frac{2π }{3} \\x=\frac{π}{6}+ k\frac{2π }{3} \end{array} \right.\)
4. cos7x-√3sin7x+√2=0
⇔$\frac{1}{2}$ cos7x-$\frac{√3}{2}$ sin7x=$\frac{√2}{7}$
⇔sin$\frac{π}{6}$cos7x-cos $\frac{π}{6}$sin7x=$\frac{-√2}{2}$
⇔sin(x-$\frac{π}{6}$ )=$\frac{-√2}{2}$
⇔7x-$\frac{π}{6}$=-$\frac{π}{4}$-k2π
⇔7x=-$\frac{π}{12}$-k2π
⇔x=-$\frac{-π}{84}$-$\frac{2π}{7}$k
