Đáp án:
\(x=\dfrac{7\pi}{12}+k.\pi\) \(k \epsilon Z\)
Giải thích các bước giải:
\(\sin (x-\dfrac{\pi}{6})=\cos (x-\dfrac{\pi}{2})\)
\(\Leftrightarrow \cos (\dfrac{\pi}{2}-(x-\dfrac{\pi}{6}))=\cos (x-\dfrac{\pi}{2})\) (hai góc phụ nhau)
\(\Leftrightarrow \cos (\dfrac{2\pi}{3}-x)=\cos (x-\dfrac{\pi}{2})\)
\(\Leftrightarrow \) \(\left[ \begin{array}{l}\dfrac{2\pi}{3}-x=x-\dfrac{\pi}{2}+k.2\pi\\\dfrac{2\pi}{3}-x=-x+\dfrac{\pi}{2}+k.2\pi\end{array} \right.\)
\(\Leftrightarrow \) \(\left[ \begin{array}{l}x =\dfrac{7\pi}{12}-k.\pi\\0x=\dfrac{\pi}{6}+ k2\pi(VN)\end{array} \right.\)
\(\Leftrightarrow x=\dfrac{7\pi}{12}+k.\pi\) \(k \epsilon Z\)
