Đáp án: $x=\dfrac{π}{15} + k \dfrac{2π}{5} ; x=k2π$
Giải thích các bước giải:
`cos(3x-π/6) - sin(2x+π/3) =0`
`<=> cos(3x- π/6) = sin(2x+π/3)`
`<=> cos(3x-π/6) = cos(π/2 - 2x - π/3)`
`<=> cos(3x-π/6) = cos(π/6 - 2x)`
`<=>` \(\left[ \begin{array}{l}3x-\dfrac{π}{6}=\dfrac{π}{6}-2x+k2π\\3x-\dfrac{π}{6} =-\dfrac{π}{6} + 2x + k2π \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{π}{15} + k \dfrac{2π}{5}\\x=k2π\end{array} \right.\)
