Đáp án+Giải thích các bước giải:
$\\$
`\qquad cos(x/2+π/6)=cos(π/2-x)`
`<=>`$\left[\begin{array}{l}\dfrac{x}{2}+\dfrac{π}{6}=\dfrac{π}{2}-x+k2π\\\dfrac{x}{2}+\dfrac{π}{6}=-(\dfrac{π}{2}-x)+k2π\end{array}\right.$`(k\in ZZ)`
`<=>`$\left[\begin{array}{l}\dfrac{3x}{2}=\dfrac{π}{3}+k2π\\\dfrac{-x}{2}=\dfrac{-2π}{3}+k2π\end{array}\right.$`(k\in ZZ)`
`<=>`$\left[\begin{array}{l}x=\dfrac{2π}{9}+\dfrac{k4π}{3}\\x=\dfrac{4π}{3}-k4π\end{array}\right.$ `(k\in ZZ)`
Vậy phương trình đã cho có họ nghiệm:
`x={2π}/9+{k4π}/3; x={4π}/3+k4π\quad (k\in ZZ)`
