1)\(cos2x+5=2\sqrt{2}\left(2-cosx\right)sin\left(x-\frac{\pi}{4}\right)\)
2)
\(sin^2x-2sinx+2=sin^23x\)
3)
\(sinx-2sin2x-sin3x=2\sqrt{2}\)
4)
\(\left(cos4x-cos2x\right)^2=5+sin3x\)
5)
\(\sqrt{5+sin^23x=sinx+2cosx}\)
6)
\(5\left(sinx+\frac{cos3x+sin3x}{1+2sin2x}\right)=cos2x+3\)
7)
\(\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right)tan\left(\frac{\pi}{4}+x\right)}=cos^44x\)
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:
\(1-t^2-4t+4=0\)
\(\Leftrightarrow t^2+4t-5=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-5\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
2.
\(\Leftrightarrow\left(sinx-1\right)^2+1=sin^23x\)
Ta có \(VT\ge1\) trong khi \(VP\le1\) với mọi x
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}sinx-1=0\\sin^23x=1\end{matrix}\right.\) \(\Leftrightarrow x=\frac{\pi}{2}+k2\pi\)
3.
\(\Leftrightarrow-2cos2x.sinx-2sin2x=2\sqrt{2}\)
\(\Leftrightarrow cos2x.sinx+sin2x=-\sqrt{2}\)
Ta có:
\(VT^2=\left(cos2x.sinx+sin2x.1\right)^2\le\left(cos^22x+sin^22x\right)\left(sin^2x+1\right)\le1\left(1+1\right)=2\)
\(\Rightarrow VT\ge-\sqrt{2}\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}sinx=1\\cos2x=sinx.sin2x\end{matrix}\right.\) (ko tồn tại x thỏa mãn)
Vậy pt vô nghiệm
4.
\(\left\{{}\begin{matrix}cos4x\le1\\-cos2x\le1\end{matrix}\right.\) \(\Rightarrow VT=\left(cos4x-cos2x\right)^2\le4\)
\(sin3x\ge-1\Rightarrow VP=5+sin3x\ge4\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left(cos4x-cos2x\right)^2=4\\sin3x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4sin^23x.sin^2x=4\\sin3x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin^2x=1\\sin3x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin^2x=1\\sinx\left(3-4sin^2x\right)=-1\end{matrix}\right.\)
\(\Leftrightarrow sinx=1\)
\(\Leftrightarrow x=\frac{\pi}{2}+k2\pi\)
5.
Chắc đề bài đúng là: \(\sqrt{5+sin^23x}=sinx+2cosx\)
Ta có: \(VT=\sqrt{5+sin^23x}\ge\sqrt{5}\)
\(VP^2=\left(1.sinx+2.cosx\right)^2\le\left(1^2+2^2\right)\left(sin^2x+cos^2x\right)=5\)
\(\Rightarrow VP\le\sqrt{5}\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}sinx;cosx>0\\sin3x=0\\2sinx=cosx\end{matrix}\right.\) (không tồn tại x thỏa mãn yêu cầu)
Vậy pt đã cho vô nghiệm
6. ĐKXĐ: \(sin2x\ne-\frac{1}{2}\)
\(5\left(sinx+\frac{4cos^3x-3cosx+3sinx-4sin^3x}{1+2sin2x}\right)=cos2x+3\)
\(\Leftrightarrow5\left(sinx+\frac{4\left(cosx-sinx\right)\left(1+sinx.cosx\right)-3\left(cosx-sinx\right)}{1+2sin2x}\right)=cos2x+3\)
\(\Leftrightarrow5\left(sinx+\frac{\left(cosx-sinx\right)\left(4+4sinx.cosx-3\right)}{1+2sin2x}\right)=cos2x+3\)
\(\Leftrightarrow5\left(sinx+\frac{\left(cosx-sinx\right)\left(1+2sin2x\right)}{1+2sin2x}\right)=cos2x+3\)
\(\Leftrightarrow5\left(sinx+cosx-sinx\right)=2cos^2x+2\)
\(\Leftrightarrow2cos^2x-5cosx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=2\left(l\right)\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\frac{\pi}{3}+k2\pi\)
