Đáp án:
`x=-\frac{π}{2}+k4π` hoặc `x=\frac{π}{6}+\frac{k4π}{3} \ (k∈\mathbb{Z})`
Giải:
Ta có:
`cosx-cos(\frac{x}{2}-\frac{π}{4})=0`
⇔ `cosx=cos(\frac{x}{2}-\frac{π}{4})`
⇔ $\left [\begin{array}{l} x=\dfrac{x}{2}-\dfrac{π}{4}+k2π \\ x=-\dfrac{x}{2}+\dfrac{π}{4}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} \dfrac{x}{2}=-\dfrac{π}{4}+k2π \\ \dfrac{3x}{2}=\dfrac{π}{4}+k2π \end{array} \right.$
⇔ $\left [\begin{array}{l} x=-\dfrac{π}{2}+k4π \\ x=\dfrac{π}{6}+\dfrac{k4π}{3} \end{array} \right. \ (k∈\mathbb{Z})$
Vậy `x=-\frac{π}{2}+k4π` hoặc `x=\frac{π}{6}+\frac{k4π}{3} \ (k∈\mathbb{Z})`
