`sin3x+sin2x=0`
⇔ `sin3x=-sin2x`
⇔ `sin3x=sin-2x`
⇔ $\left [\begin{array}{l} 3x=-2x+k2π \\ 3x=π+2x+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} 5x=k2π \\ x=π+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{k2π}{5} \\ x=π+k2π \end{array} \right. \ (k∈\mathbb{Z})$
`sinx=cos3x`
⇔ `sinx=sin(\frac{π}{2}-3x)`
⇔ $\left [\begin{array}{l} x=\dfrac{\pi}{2}-3x+k2π \\ x=π-\dfrac{π}{2}+3x+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} 4x=\dfrac{\pi}{2}+k2π \\ -2x=\dfrac{\pi}{2}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{\pi}{8}+\dfrac{kπ}{2} \\ x=-\dfrac{\pi}{4}-kπ \end{array} \right. \ (k∈\mathbb{Z})$
`sin4x+cosx=0`
⇔ `sin4x=-cosx`
⇔ `sin4x=cos(π-x)`
⇔ `sin4x=sin(x-\frac{π}{2})`
⇔ $\left [\begin{array}{l} 4x=x-\dfrac{π}{2}+k2π \\ 4x=π-x+\dfrac{π}{2}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} 3x=-\dfrac{π}{2}+k2π \\ 5x=\dfrac{3π}{2}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x=-\dfrac{π}{6}+\dfrac{k2π}{3} \\ x=\dfrac{3π}{10}+\dfrac{k2π}{5} \end{array} \right. \ (k∈\mathbb{Z})$
