Hỏi lm j?

\(\Leftrightarrow\dfrac{\sqrt{2}}{2}\sin x+\dfrac{\sqrt{2}}{2}\cos x=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\sin x\cdot\cos\dfrac{\pi}{4}+\cos x\cdot\sin\dfrac{\pi}{4}=\dfrac{\pi}{4}\)

\(\Leftrightarrow\sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

D

a \(Vì\) \(-1\le\cos2x\le1\)

\(\Leftrightarrow-7\le7\cos2x\le7\)

\(\Leftrightarrow-10\le3+7\cos2x\le4\)

\(Vậy\)  \(y_{max}=4\)

        \(y_{min}=-10\)

b \(Vì\) \(-1\le\sin3x\le1\)

\(\Leftrightarrow5\ge-5\sin3x\ge-5\)

\(\Leftrightarrow7\ge2-5\sin3x\ge-3\)

\(\Leftrightarrow-3\le2-5\sin3x\le7\)

\(Vậy\) \(y_{max}=7\)

        \(y_{min}=-3\)

Cần lời giải ko bạn :> ?

dấu tương đương tứ 2 thiếu \(k2\pi\) ở hàng dưới

mk tính luôn nên hơi tắt chút xíu

\(\Leftrightarrow\sin\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\2x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=k2\pi\\2x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\left(k\in Z\right)\)

\(\Leftrightarrow\cos\left(2x-10^o\right)=\cos120^o\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-10^o=120^o+k360^o\\2x-10^o=180^o-120^o+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=130^o+k360^o\\2x=70^o+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=65^o+k180^o\\x=35^o+k180^o\end{matrix}\right.\left(k\in Z\right)\)