Question

I don't Know có lm đc nổi câu này hay ko? >:)

Tìm số nghiệm phương trình: \(\sin\left(2x-\dfrac{\pi}{6}\right)=1\) trên \(\left[\pi;2\pi\right]\)

Answer 1

\(\sin\left(2x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow2x=\dfrac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\left(k\in Z\right)\)

\(Vì\) \(x\in\left[\pi;2\pi\right]\) ta có:

\(\pi\le\dfrac{\pi}{3}+k\pi\le2\pi\)

\(\Leftrightarrow\dfrac{2\pi}{3}\le k\pi\le\dfrac{5\pi}{3}\)

\(\Leftrightarrow\dfrac{2}{3}\le k\le\dfrac{5}{3}\)

\(\Leftrightarrow0.7\approx\dfrac{2}{3}\le k\le\dfrac{5}{3}\approx1.7\)

Do \(k\in Z\) nên k = 1

Vậy PT có 1 nghiệm / \(\left[\pi;2\pi\right]\)