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6.
\(\Leftrightarrow\frac{1}{2}cos6x+\frac{1}{2}cos4x=\frac{1}{2}cos6x+\frac{1}{2}cos2x+\frac{3}{2}+\frac{3}{2}cos2x+1\)
\(\Leftrightarrow cos4x=4cos2x+5\)
\(\Leftrightarrow2cos^22x-1=4cos2x+5\)
\(\Leftrightarrow cos^22x-2cos2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3>1\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
7.
Thay lần lượt 4 đáp án ta thấy chỉ có đáp án C thỏa mãn
8.
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=\left\{\frac{\pi}{6};\frac{\pi}{2}\right\}\)
9.
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}-1\le t\le1\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Rightarrow mt+\frac{t^2-1}{2}+1=0\)
\(\Leftrightarrow t^2+2mt+1=0\)
Pt đã cho có đúng 1 nghiệm thuộc \(\left[-1;1\right]\) khi và chỉ khi: \(\left[{}\begin{matrix}m\ge1\\m\le-1\end{matrix}\right.\)
10.
\(\frac{\sqrt{3}}{2}cos5x-\frac{1}{2}sin5x=cos3x\)
\(\Leftrightarrow cos\left(5x-\frac{\pi}{6}\right)=cos3x\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\frac{\pi}{6}=3x+k2\pi\\5x-\frac{\pi}{6}=-3x+k2\pi\end{matrix}\right.\)
7.
Đặt \(\left|sinx+cosx\right|=\left|\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\right|=t\Rightarrow0\le t\le\sqrt{2}\)
Ta có: \(t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\frac{t^2-1}{2}\) (1)
Pt trở thành:
\(\frac{t^2-1}{2}+t=1\)
\(\Leftrightarrow t^2+2t-3=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
Thay vào (1) \(\Rightarrow2sinx.cosx=t^2-1=0\)
\(\Leftrightarrow sin2x=0\Rightarrow x=\frac{k\pi}{2}\)
\(\Rightarrow x=\left\{\frac{\pi}{2};\pi;\frac{3\pi}{2}\right\}\Rightarrow\sum x=3\pi\)
6.
\(\Leftrightarrow\left(1-sin2x\right)+sinx-cosx=0\)
\(\Leftrightarrow\left(sin^2x+cos^2x-2sinx.cosx\right)+sinx-cosx=0\)
\(\Leftrightarrow\left(sinx-cosx\right)^2+sinx-cosx=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx-cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\sinx-cosx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=k\pi\\x-\frac{\pi}{4}=-\frac{\pi}{4}+k\pi\\x-\frac{\pi}{4}=\frac{5\pi}{4}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=k\pi\\x=\frac{3\pi}{2}+k\pi\end{matrix}\right.\)
Pt có 3 nghiệm trên đoạn đã cho: \(x=\left\{\frac{\pi}{4};0;\frac{\pi}{2}\right\}\)
a/
Đặt \(cosx=t\Rightarrow0< t\le1\)
\(\Rightarrow t^2-2mt+4\left(m-1\right)=0\)
\(\Leftrightarrow t^2-4-2m\left(t-2\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+2-2m\right)=0\)
\(\Leftrightarrow t=2m-2\)
\(\Rightarrow0< 2m-2\le1\Rightarrow1< m\le\frac{3}{2}\)
b.
\(x\in\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\Rightarrow\frac{x}{2}\in\left(-\frac{\pi}{4};\frac{\pi}{4}\right)\)
Đặt \(sin\frac{x}{2}=t\Rightarrow-\frac{\sqrt{2}}{2}< t< \frac{\sqrt{2}}{2}\)
\(\Rightarrow4t^2+2t+m-2=0\Leftrightarrow4t^2+2t-2=-m\)
Xét \(f\left(t\right)=4t^2+2t-2\) trên \(\left(-\frac{\sqrt{2}}{2};\frac{\sqrt{2}}{2}\right)\)
\(f\left(-\frac{\sqrt{2}}{2}\right)=-\sqrt{2}\) ; \(f\left(\frac{\sqrt{2}}{2}\right)=\sqrt{2}\) ; \(f\left(-\frac{1}{4}\right)=-\frac{9}{4}\)
\(\Rightarrow-\frac{9}{4}\le f\left(t\right)< \sqrt{2}\Rightarrow-\frac{9}{4}\le-m< \sqrt{2}\)
\(\Rightarrow-\sqrt{2}< m\le\frac{9}{4}\)
\(\sqrt{3}sin2x-\left(1+cos2x\right)=m\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=\frac{m+1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=\frac{m+1}{2}\)
Do \(x\in\left[\frac{\pi}{4};\frac{5\pi}{12}\right]\Rightarrow2x-\frac{\pi}{6}\in\left[\frac{\pi}{3};\frac{2\pi}{3}\right]\)
\(\Rightarrow sin\left(2x-\frac{\pi}{6}\right)\in\left[\frac{\sqrt{3}}{2};1\right]\)
Pt có nghiệm thuộc đoạn đã cho khi và chỉ khi: \(\frac{\sqrt{3}}{2}\le\frac{m+1}{2}\le1\)
\(\Leftrightarrow\sqrt{3}-1\le m\le1\)
\(\Leftrightarrow2x+\frac{\pi}{3}=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{2}\)
Do \(x\in\left[0;2\pi\right]\Rightarrow0\le\frac{\pi}{12}+\frac{k\pi}{2}\le2\pi\)
\(\Rightarrow-\frac{1}{6}\le k\le\frac{23}{6}\Rightarrow k=\left\{0;1;2;3\right\}\)
\(\Rightarrow x=\left\{\frac{\pi}{12};\frac{7\pi}{12};\frac{13\pi}{12};\frac{19\pi}{12}\right\}\)
\(\Leftrightarrow cosx-2cosx+m=0\)
\(\Leftrightarrow cosx=-m\)
Từ đường tròn lượng giác ta thấy để pt có đúng 3 nghiệm pb thuộc \(\left[0;\frac{7\pi}{2}\right]\Rightarrow0\le-m< 1\Rightarrow-1< m\le0\)
c/
\(\Leftrightarrow cos^3\left(x-\frac{\pi}{3}\right)=\frac{1}{8}\)
\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=k2\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{6}-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{\pi}{6}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{6}+\frac{k2\pi}{3}\)
b/
\(\Rightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow-cos2x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos2x=-sin\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{5\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=x+\frac{5\pi}{6}+k2\pi\\2x=-x-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}+k2\pi\\x=-\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)