\(Ta\) \(có\): \(\sqrt{a^2+b^2}=\sqrt{\sqrt{3}^2+1^2}=2\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}\sin x+\dfrac{1}{2}\cos x=\dfrac{1}{2}\)

\(\Leftrightarrow\sin x\cdot\cos\dfrac{\pi}{6}+\cos x\cdot\sin\dfrac{\pi}{6}=\dfrac{1}{2}\)

\(\Leftrightarrow\sin\left(x+\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\x+\dfrac{\pi}{6}=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)

bây h đang ôn để mai kiểm tra r

MÃ CÓ 4 ĐỀ AHHHH

REFER

Đất là môi trường, là nền cho cây trồng phát triển. Đất cung cấp nước, oxi, dinh dưỡng và giúp cây đứng vững, không bị đổ. Nhờ đó cây trồng có thể phát triển tốt và cho ra sản phẩm phục vụ đời sống con người.

sai r Huy ơi

\(\Leftrightarrow1-\cos^2x+\cos x-1=0\)

\(\Leftrightarrow-\cos^2x+\cos x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos x=0\\\cos=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=k2\pi\end{matrix}\right.\left(k\in Z\right)\)

\(a\) \(\cot\left(3x-\dfrac{\pi}{3}\right)=\dfrac{1}{\sqrt{3}}\)

\(\Leftrightarrow3x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow3x=\dfrac{2\pi}{3}+k\pi\)

\(\Leftrightarrow x=\dfrac{2\pi}{9}+k\dfrac{\pi}{3}\left(k\in Z\right)\)

\(b\) \(2\cos\left(3x-\dfrac{\pi}{4}\right)+1=0\)

\(\Leftrightarrow\cos\left(3x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow\cos\left(3x-\dfrac{\pi}{4}\right)=\cos\dfrac{\pi}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{4}=-\dfrac{\pi}{3}+k2\pi\\3x-\dfrac{\pi}{4}=\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{-\pi}{12}+k2\pi\\3x=\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{36}+k\dfrac{2\pi}{3}\\x=\dfrac{7\pi}{36}+k\dfrac{2\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)

a \(\sin\left(2x-\dfrac{\pi}{3}\right)=1\)

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

\(\Leftrightarrow2x=\dfrac{5\pi}{6}+k2\pi\)

\(\Leftrightarrow x=\dfrac{5\pi}{12}+k\pi\left(k\in Z\right)\)

b \(\tan\left(2x\right)=\sqrt{3}\)

\(\Leftrightarrow2x=\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k\dfrac{\pi}{2}\left(k\in Z\right)\)