Đáp án:
`S={\frac{π}{2}+kπ, arcsin\frac{1}{4}+k2π, π-arcsin\frac{1}{4}+k2π | k∈\mathbb{Z}}`
Giải:
`cosx-2sin2x=0`
⇔ `cosx-4sinxcosx=0`
⇔ `cosx(1-4sinx)=0`
⇔ $\left [\begin{array}{l} cosx=0 \\ 1-4sinx=0 \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{\pi}{2}+kπ \\ 4sinx=1 \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{\pi}{2}+kπ \\ sinx=\dfrac{1}{4} \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{\pi}{2}+kπ \\ x=arcsin\dfrac{1}{4}+k2π \\ x=π-arcsin\dfrac{1}{4}+k2π \end{array} \right. \ (k∈\mathbb{Z})$
Vậy `S={\frac{π}{2}+kπ, arcsin\frac{1}{4}+k2π, π-arcsin\frac{1}{4}+k2π | k∈\mathbb{Z}}`
