giải phương trình
1.\(sin^3x+2cosx-2+sin^2x=0\)
\(2.\frac{\sqrt{3}}{2}sin2x+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
3.\(2sin2x-cos2x=7sinx+2cosx-4\)
4.\(2cos2x-8cosx+7=\frac{1}{cosx}\)
5.\(cos^8x+sin^8x=2\left(cos^{10}x+sin^{10}x\right)+\frac{5}{4}cos2x\)
6.\(1+sinx+cos3x=cosx+sin2x+cos2x\)
7.\(1+sinx+cosx+sin2x+cos2x=0\)
1.
\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Leftrightarrow...\\sinx+cosx+sinx.cosx-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
2.
\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)
Xét (1):
Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm
3.
\(\Leftrightarrow4sinx.cosx-\left(1-2sin^2x\right)=7sinx+2cosx-4\)
\(\Leftrightarrow2cosx\left(2sinx-1\right)+2sin^2x-7sinx+3=0\)
\(\Leftrightarrow2cosx\left(2sinx-1\right)+\left(sinx-3\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2cosx+sinx-3\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\Leftrightarrow...\\2cosx+sinx=3\left(1\right)\end{matrix}\right.\)
Xét (1), do \(2^2+1^2< 3^2\) nên (1) vô nghiệm
4.
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow2cosx\left(2cos^2x-1\right)-8cos^2x+7cosx=1\)
\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)
\(\Leftrightarrow\left(cosx-1\right)\left(2cosx-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
5.
\(\Leftrightarrow2cos^{10}x-cos^8x+2sin^{10}x-sin^8x+\frac{5}{4}cos2x=0\)
\(\Leftrightarrow cos^8x\left(2cos^2x-1\right)-sin^8x\left(1-2sin^2x\right)+\frac{5}{4}cos2x=0\)
\(\Leftrightarrow cos2x.cos^8x-cos2x.sin^8x+\frac{5}{4}cos2x=0\)
\(\Leftrightarrow cos2x\left(cos^8x-sin^8x+\frac{5}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\Leftrightarrow...\\cos^8x-sin^8x+\frac{5}{4}=0\left(1\right)\end{matrix}\right.\)
Xét (1);
\(\Leftrightarrow\left(sin^4x+cos^4x\right)\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+\frac{5}{4}=0\)
\(\Leftrightarrow\left(1-\frac{1}{2}sin^22x\right)cos2x+\frac{5}{4}=0\)
\(\Leftrightarrow\left(1-\frac{1}{2}\left(1-cos^22x\right)\right)cos2x+\frac{5}{4}=0\)
\(\Leftrightarrow2cos^3x+2cos2x+5=0\)
Pt này nghiệm rất xấu
6.
\(\Leftrightarrow1+sinx-sin2x+cos3x-cosx-\left(1-2sin^2x\right)=0\)
\(\Leftrightarrow sinx-sin2x-2sin2x.sinx+2sin^2x=0\)
\(\Leftrightarrow sinx-sin2x+2sinx\left(sinx-sin2x\right)=0\)
\(\Leftrightarrow\left(sinx-sin2x\right)\left(1+sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=sinx\\sinx=-\frac{1}{2}\\\end{matrix}\right.\)
\(\Leftrightarrow...\)
7.
\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)
\(\Leftrightarrow sinx+cosx+2cosx\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow...\)
