a) `sin(x-\frac{π}{5})=sin\frac{π}{5}`
⇔ $\left [\begin{array}{l} x-\dfrac{π}{5}=\dfrac{π}{5}+k2π \\ x-\dfrac{π}{5}=π-\dfrac{π}{5}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x-\dfrac{π}{5}=\dfrac{π}{5}+k2π \\ x-\dfrac{π}{5}=\dfrac{4π}{5}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{2π}{5}+k2π \\ x=π+k2π \end{array} \right. \ (k∈\mathbb{Z})$
b) `sin3x=sin(x+\frac{π}{6})`
⇔ $\left [\begin{array}{l} 3x=x+\dfrac{π}{6}+k2π \\ 3x=π-x-\dfrac{π}{6}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} 2x=\dfrac{π}{6}+k2π \\ 4x=\dfrac{5π}{6}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{π}{12}+kπ \\ x=\dfrac{5π}{24}+\dfrac{kπ}{2} \end{array} \right. \ (k∈\mathbb{Z})$
c) `sin3x=\frac{\sqrt{2}}{2}`
⇔ `sin3x=sin\frac{π}{4}`
⇔ $\left [\begin{array}{l} 3x=\dfrac{π}{4}+k2π \\ 3x=π-\dfrac{π}{4}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} 3x=\dfrac{π}{4}+k2π \\ 3x=\dfrac{3π}{4}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{π}{12}+\dfrac{k2π}{3} \\ x=\dfrac{π}{4}+\dfrac{k2π}{3} \end{array} \right. \ (k∈\mathbb{Z})$
