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1/ ĐKXĐ: \(\cos2x\ne0\)
\(\frac{\cos4x}{\cos2x}=\frac{\sin2x}{\cos2x}\)\(\Leftrightarrow\cos4x-\sin2x=0\)
\(\Leftrightarrow2\cos^22x-1-\sin2x=0\)
\(\Leftrightarrow2-2\sin^22x-1-\sin2x=0\)
\(\Leftrightarrow2\sin^22x+\sin2x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\frac{1}{2}=\sin\frac{\pi}{6}\\\sin2x=-1=\sin\frac{-\pi}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{6}+2k\pi\\2x=\frac{5\pi}{6}+2k\pi\\2x=\frac{-\pi}{2}+2k\pi\left(l\right)\\2x=\frac{3\pi}{2}+2k\pi\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)
2/ \(\sin2.4x+\cos4x=1+2\sin2x.\cos\left(2x+4x\right)\)
\(\Leftrightarrow2\sin4x.\cos4x+\cos4x=1+2\sin2x.\left(\cos2x.\cos4x-\sin2x.\sin4x\right)\)
\(\Leftrightarrow2\sin4x.\cos4x+\cos4x=1+2\sin2x.\cos2x.\cos4x-2\sin^22x.\sin4x\)
\(\Leftrightarrow2\sin4x.\cos4x+\cos4x=1+\sin4x.\cos4x-\sin4x+\cos4x.\sin4x\)
Đến đây bn tự giải nốt nhé, lm kiểu bthg thôi bởi vì đã quy về hết sin4x và cos4x r
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.
\(\frac{\sqrt{3}}{2}cos5x-\frac{1}{2}sin5x=cos3x\)
\(\Leftrightarrow cos\left(5x-\frac{\pi}{6}\right)=cos3x\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\frac{\pi}{6}=3x+k2\pi\\5x-\frac{\pi}{6}=-3x+k2\pi\end{matrix}\right.\)
d/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+4=4\left(\sqrt{3}sinx+cosx\right)\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+\frac{5}{2}=4\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow2sin^2\left(x+\frac{\pi}{6}\right)+4sin\left(x+\frac{\pi}{6}\right)-\frac{7}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{-2+\sqrt{11}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{-2-\sqrt{11}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\\x=\frac{5\pi}{6}-arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
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: \(x=\left\{\frac{\pi}{4};0;\frac{\pi}{2}\right\}\)
ngại viết quá hihi, mà hơi ngáo tí cái dạng này lm rồi mà cứ quên
bài trước mk bình luận bạn đọc chưa nhỉ
7.
ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(\frac{\pi}{4}-x\right).sin\left(\frac{\pi}{4}+x\right)\ne0\\cos\left(\frac{\pi}{4}-x\right)cos\left(\frac{\pi}{4}+x\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow cos2x\ne0\)
Phương trình tương đương:
\(\Leftrightarrow\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right).cot\left(\frac{\pi}{2}-\frac{\pi}{4}-x\right)}=cos^44x\)
\(\Leftrightarrow\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right).cot\left(\frac{\pi}{4}-x\right)}=cos^24x\)
\(\Leftrightarrow sin^42x+cos^42x=cos^44x\)
\(\Leftrightarrow\left(sin^22x+cos^22x\right)^2-2sin^22x.cos^22x=cos^44x\)
\(\Leftrightarrow1-\frac{1}{2}sin^24x=cos^44x\)
\(\Leftrightarrow2-\left(1-cos^24x\right)=2cos^44x\)
\(\Leftrightarrow2cos^44x-cos^24x-1=0\)
\(\Leftrightarrow\left(cos^24x-1\right)\left(2cos^24x+1\right)=0\)
\(\Leftrightarrow cos^24x-1=0\)
\(\Leftrightarrow sin^24x=0\Leftrightarrow sin4x=0\)
\(\Leftrightarrow2sin2x.cos2x=0\Leftrightarrow sin2x=0\)
\(\Leftrightarrow x=\frac{k\pi}{2}\)
1.
\(cos2x+5=2\left(2-cosx\right)\left(sinx-cosx\right)\)
\(\Leftrightarrow2cos^2x+4=4sinx-4cosx-2sinx.cosx+2cos^2x\)
\(\Leftrightarrow2sinx.cosx-4\left(sinx-cosx\right)+4=0\)
Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=1-t^2\end{matrix}\right.\)
Pt trở thành:
\(1-t^2-4t+4=0\)
\(\Leftrightarrow t^2+4t-5=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-5\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
