giải PT sau:
\(\sin^2x+\cos x-1=0\)
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\(a\) \(\cot\left(3x-\dfrac{\pi}{3}\right)=\dfrac{1}{\sqrt{3}}\)
\(\Leftrightarrow3x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k\pi\)
\(\Leftrightarrow3x=\dfrac{2\pi}{3}+k\pi\)
\(\Leftrightarrow x=\dfrac{2\pi}{9}+k\dfrac{\pi}{3}\left(k\in Z\right)\)
\(b\) \(2\cos\left(3x-\dfrac{\pi}{4}\right)+1=0\)
\(\Leftrightarrow\cos\left(3x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)
\(\Leftrightarrow\cos\left(3x-\dfrac{\pi}{4}\right)=\cos\dfrac{\pi}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{4}=-\dfrac{\pi}{3}+k2\pi\\3x-\dfrac{\pi}{4}=\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{-\pi}{12}+k2\pi\\3x=\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{36}+k\dfrac{2\pi}{3}\\x=\dfrac{7\pi}{36}+k\dfrac{2\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)
a: =>2x-pi/3=pi/2+k2pi
=>x=5/12pi+kpi
b: =>2x=pi/3+kpi
=>x=pi/6+kpi/2
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 \(\tan\left(2x\right)=\sqrt{3}\)
\(\Leftrightarrow2x=\dfrac{\pi}{3}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{6}+k\dfrac{\pi}{2}\left(k\in Z\right)\)
\(\Leftrightarrow\dfrac{\sqrt{2}}{2}\sin x+\dfrac{\sqrt{2}}{2}\cos x=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sin x\cdot\cos\dfrac{\pi}{4}+\cos x\cdot\sin\dfrac{\pi}{4}=\dfrac{\pi}{4}\)
\(\Leftrightarrow\sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
Hàm số đạt giá trị nhỏ nhất tại M(2; - 3)
⇒ \(\left\{{}\begin{matrix}\dfrac{-b}{2a}=2\\4a+2b+1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4a+b=0\\4a+2b=-4\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}a=1\\b=-4\end{matrix}\right.\)
Vậy b = -4
Đặt độ dài cạnh của hình lập phương là a
Ta có : \(\overrightarrow{AG}.\overrightarrow{BE}=\left(\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AE}\right)\left(\overrightarrow{AE}-\overrightarrow{AB}\right)\)
⇒ \(\overrightarrow{AG}.\overrightarrow{BE}=\overrightarrow{AB}.\overrightarrow{AE}-AB^2+\overrightarrow{AD}.\overrightarrow{AE}-\overrightarrow{AD}.\overrightarrow{AB}+AE^2-\overrightarrow{AE}.\overrightarrow{AB}\)
Các tích vô hướng ở vế phải bằng 0 và AE = AB
⇒ \(\overrightarrow{AG}.\overrightarrow{BE}=0\) ⇒ AG ⊥ BE, tức góc giữa hai đường thẳng này bằng 900
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 \(Vì\) \(-1\le\sin3x\le1\)
\(\Leftrightarrow5\ge-5\sin3x\ge-5\)
\(\Leftrightarrow7\ge2-5\sin3x\ge-3\)
\(\Leftrightarrow-3\le2-5\sin3x\le7\)
\(Vậy\) \(y_{max}=7\)
\(y_{min}=-3\)
tham khảo
\(\Leftrightarrow1-\cos^2x+\cos x-1=0\)
\(\Leftrightarrow-\cos^2x+\cos x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\cos x=0\\\cos=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=k2\pi\end{matrix}\right.\left(k\in Z\right)\)