Đáp án: `x=\dfrac{-π}{4} + k π`
Giải thích các bước giải:
ĐK: sin(x-π/4) ne 0 <=> x - \pi/4 ne k\pi <=> x ne \pi/4+k\pi`
` cos2x . cot(x-π/4) = 0`
`<=>` \(\left[ \begin{array}{l}cos2x=0\\cot(x-\dfrac{π}{4} ) =0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=\dfrac{π}{2} + kπ\\x-\dfrac{π}{4}=\dfrac{π}{2} + kπ\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{π}{4} + k \dfrac{π}{2}\\x=\dfrac{3π}{4} + kπ\end{array} \right.\)
Vậy `x=-\pi/4+kπ`.
