Bài 4: Ôn tập chương Hàm số lượng giác và phương trình lượng giác - Hỏi đáp

\(\Leftrightarrow cosx-cos7x-3\sqrt{3}sinx=0\)

\(\Leftrightarrow2sin4x.sin3x-3\sqrt{3}sinx=0\)

\(\Leftrightarrow2sin4x.\left(3sinx-4sin^3x\right)-3\sqrt{3}sinx=0\)

\(\Leftrightarrow sinx\left(6sin4x-8sin^2x.sin4x-3\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=...\\6sin4x-8sin^2x.sin4x=3\sqrt{3}\left(1\right)\end{matrix}\right.\)

Xét \(\left(1\right)\Leftrightarrow6sin4x-4sin4x\left(1-cos2x\right)=3\sqrt{3}\)

\(\Leftrightarrow2sin4x+4sin4x.cos2x=3\sqrt{3}\)

\(\Leftrightarrow sin4x+4sin2x.cos^22x=\frac{3\sqrt{3}}{2}\)

Ta có:

\(1=sin^22x+\frac{cos^22x}{2}+\frac{cos^22x}{2}\ge3\sqrt[3]{\frac{\left(sin2x.cos^22x\right)^2}{4}}\)

\(\Rightarrow\left(sin2x.cos^22x\right)^2\le\frac{4}{27}\Rightarrow sin2x.cos^22x\le\frac{2\sqrt{3}}{9}\)

\(\Rightarrow sin4x+4sin2x.cos^22x\le1+\frac{8\sqrt{3}}{9}< \frac{3\sqrt{3}}{2}\) nên pt vô nghiệm

\(2sin\left(x+4\pi\right)=\frac{2m-1}{2}\)

\(\Rightarrow-1\le\frac{2m-1}{2}\le1\Rightarrow\frac{-1}{2}\le m\le\frac{3}{2}\)

\(sin\left(7-x\right)=5m+6m^2\)

\(\Rightarrow\left\{{}\begin{matrix}6m^2+5m\ge-1\\6m^2+5m\le1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6m^2+5m+1\ge0\\6m^2+5m-1\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m\ge-\frac{1}{3}\\m\le-\frac{1}{2}\end{matrix}\right.\\-1\le m\le\frac{1}{6}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le m\le-\frac{1}{2}\\-\frac{1}{3}\le m\le\frac{1}{6}\end{matrix}\right.\)

a) Ta có : \(sin\left(x-\frac{2\pi}{3}\right)=cos2x\)

\(\Leftrightarrow sin\left(x-\frac{2\pi}{3}\right)=sin\left(\frac{\pi}{2}-2x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{18}+k\frac{2\pi}{3}\\x=-\frac{7\pi}{6}-k2\pi\end{matrix}\right.\)

Vậy ...

\(\Leftrightarrow\left(2\sin x+1\right)\left(\sqrt{3}\sin x+2\cos^2x-1\right)-\sin2x-\cos x=0\Leftrightarrow\left(2\sin x+1\right)\left(\sqrt{3}\sin x+2\cos^2x-1-2\cos^2x+1-\cos x\right)=0\Leftrightarrow\left(2\sin x+1\right)\left(\sqrt{3}\sin x-\cos x\right)=0\Rightarrow\left[{}\begin{matrix}2\sin x+1=0\\\sqrt{3}\sin x-\cos x=0\end{matrix}\right.\)

Đặt \(a=3sinx-4cosx\Rightarrow a^2\le\left(3^2+4^2\right)\left(sin^2x+cos^2x\right)=25\)

\(\Rightarrow-5\le a\le5\)

\(y=a^2-2a+1\ge2m\)

\(\Rightarrow\left(a-1\right)^2\ge2m\)

Để BPT đúng với mọi x thuộc R

\(\Leftrightarrow2m\le\min\limits_{\left[-5;5\right]}\left(a-1\right)^2\)

Mà \(\left(a-1\right)^2\ge0\) \(\forall a\Rightarrow2m\le0\Rightarrow m\le0\)

\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=cos2x\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos2x\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)