Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC

DN

Mn giúp em giải và giải thích từng câu với ạ

NL
15 tháng 7 2021 lúc 23:49

13.

\(y=1+sin2x-\left(1-sin^22x\right)=sin^22x+sin2x\)

\(y=\left(sin2x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(\left\{{}\begin{matrix}sin^22x\le1\\sin2x\le1\end{matrix}\right.\) \(\Rightarrow y\le1+1=2\)

\(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{4}\\b=2\end{matrix}\right.\) 

\(\Rightarrow4a+b=1\)

14.

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=x+\dfrac{3\pi}{4}+k2\pi\\2x-\dfrac{\pi}{4}=\dfrac{\pi}{4}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

\(\Rightarrow x=\left\{\dfrac{\pi}{6};\dfrac{5\pi}{6}\right\}\)

\(\Rightarrow\dfrac{\pi}{6}+\dfrac{5\pi}{6}=\pi\)

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NL
15 tháng 7 2021 lúc 23:55

15.

\(3cosx+2cos^2x-1-cos3x+1=cosx-cos3x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=-1\end{matrix}\right.\)

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

\(\Rightarrow\) Nghiệm lớn nhất \(x=\dfrac{3\pi}{2}\)

\(sin\left(\dfrac{3\pi}{2}-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)

16.

\(cos\left(2x+\dfrac{2\pi}{3}\right)+4cos\left(\dfrac{\pi}{6}-x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow cos\left[\pi-2\left(\dfrac{\pi}{6}-x\right)\right]+4cos\left(\dfrac{\pi}{6}-x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow-cos\left[2\left(\dfrac{\pi}{6}-x\right)\right]+4cos\left(\dfrac{\pi}{6}-x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow1-2cos^2\left(\dfrac{\pi}{6}-x\right)+4cos\left(\dfrac{\pi}{6}-x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow1-2t^2+4t=\dfrac{5}{2}\Leftrightarrow4t^2-8t+3=0\)

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