cos2x + cosx + 1 = 0
a, cos3x + cos2x - cosx - 1 = 0
b, cos(8sinx) = 1
c, 1 + cos2x + cosx = 0
d, 3cosx + |sinx| = 2
a/
\(\Leftrightarrow4cos^3x-3cosx+2cos^2x-1-cosx-1=0\)
\(\Leftrightarrow2cos^3x+cos^2x-2cosx-1=0\)
\(\Leftrightarrow cos^2x\left(2cosx+1\right)-\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left(cos^2x-1\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\2cosx+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
b/
\(cos\left(8sinx\right)=1\)
\(\Leftrightarrow8sinx=k2\pi\)
\(\Leftrightarrow sinx=\frac{k\pi}{4}\)
Do \(-1\le sinx\le1\Rightarrow-1\le\frac{k\pi}{4}\le1\)
\(\Rightarrow k=\left\{-1;0;1\right\}\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{\pi}{4}\\sinx=0\\sinx=\frac{\pi}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm arcsin\left(\frac{\pi}{4}\right)+k2\pi\\x=\pi\pm arcsin\left(\frac{\pi}{4}\right)+k2\pi\\x=k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow1+2cos^2x-1+cosx=0\)
\(\Leftrightarrow2cos^2x-cosx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
d/
Đặt \(\left\{{}\begin{matrix}\left|sinx\right|=a\ge0\\cosx=b\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}a+3b=2\\a^2+b^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2-3b\\a^2+b^2=1\end{matrix}\right.\)
\(\Rightarrow\left(2-3b\right)^2+b^2-1=0\)
\(\Rightarrow10b^2-12b+3=0\Rightarrow\left[{}\begin{matrix}b=\frac{6+\sqrt{6}}{10}\Rightarrow a=\frac{2-3\sqrt{6}}{10}\left(l\right)\\b=\frac{6-\sqrt{6}}{10}\Rightarrow a=\frac{2+3\sqrt{6}}{10}\end{matrix}\right.\)
\(\Rightarrow cosx=\frac{6-\sqrt{6}}{10}\)
\(\Rightarrow x=\pm arccos\left(\frac{6-\sqrt{6}}{10}\right)+k2\pi\)
\(sinx+4cosx=2+sin2x\)
\(\left(1-sin2x\right)\left(sinx+cosx\right)=cos2x\)
\(1+sinx+cosx+sin2x+cos2x=0\)
\(sinx+sin2x+sin3x=1+cosx+cos2x\)
\(sin^22x-cos^28x=sin\left(\dfrac{17\pi}{2}+10x\right)\)
Giải pt
\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)
\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)
\(sin2x-cos2x+3sinx-cosx-1=0\)
1.
\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)
\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)
\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)
\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)
Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)
\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm:
\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)
2.
\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)
\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)
\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
giải các pt sau:
a) cosx(1-cos2x) - sin^2x = 0
b) sin3x + cos2x = 1 + 2sinxcos3x
c) ( cosx+1)(sinx - cosx + 3) = sin^2x
d) (1+sinx)(cosx-sinx) = cos^2x
a.
\(\Leftrightarrow cosx\left[1-\left(1-2sin^2x\right)\right]-sin^2x=0\)
\(\Leftrightarrow2sin^2x.cosx-sin^2x=0\)
\(\Leftrightarrow sin^2x\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
b.
Câu b chắc chắn đề đúng chứ bạn? Vế phải ấy?
c/
\(\left(1+cosx\right)\left(sinx-cosx+3\right)=1-cos^2x\)
\(\Leftrightarrow\left(1+cosx\right)\left(sinx-cosx+3\right)-\left(1+cosx\right)\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1+cosx\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow cosx=-1\)
\(\Leftrightarrow x=\pi+k2\pi\)
d.
\(\Leftrightarrow\left(1+sinx\right)\left(cosx-sinx\right)=1-sin^2x\)
\(\Leftrightarrow\left(1+sinx\right)\left(cosx-sinx\right)-\left(1+sinx\right)\left(1-sinx\right)=0\)
\(\Leftrightarrow\left(1+sinx\right)\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=k2\pi\end{matrix}\right.\)
1) So nghiem phuong trinh \(\dfrac{\left(1+cos2x+sin2x\right)cosx+cos2x}{1+tanx}=cosx\) voi x ∈ (0; \(\dfrac{\Pi}{2}\)) la: (giai ra nua nha)
A. 0 B. 1 C. 2 D. 3
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+k\pi\\x\ne-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{\left(1+2cos^2x-1+2sinx.cosx\right)cosx+cos^2x-sin^2x}{1+\dfrac{sinx}{cosx}}=cosx\)
\(\Leftrightarrow\dfrac{2cos^2x\left(sinx+cosx\right)+\left(sinx+cosx\right)\left(cosx-sinx\right)}{\dfrac{sinx+cosx}{cosx}}=cosx\)
\(\Leftrightarrow\dfrac{cosx\left(sinx+cosx\right)\left(2cos^2x+cosx-sinx\right)}{sinx+cosx}=cosx\)
\(\Rightarrow2cos^2x+cosx-sinx=1\)
\(\Rightarrow cosx-sinx-cos2x=0\)
\(\Rightarrow cosx-sinx-\left(cos^2x-sin^2x\right)=0\)
\(\Rightarrow cosx-sinx-\left(cosx-sinx\right)\left(cosx+sinx\right)=0\)
\(\Rightarrow\left(cosx-sinx\right)\left(1-sinx-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=\dfrac{\pi}{4}\)
Có 1 nghiệm trên khoảng đã cho
a)căn 3 sin4x-cos4x-2cosx=0
b)cosx +căn 3 cos2x-căn 3 sinx-sin2x=0
c)cos 3x+sin2x=căn 3(sin3x+cos2x)
d)cosx +căn 3=3-3/cosx+căn 3 sinx+1
a/
\(\sqrt{3}sin4x-cos4x=2cosx\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x-\frac{1}{2}cos4x=cosx\)
\(\Leftrightarrow sin\left(4x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{6}=\frac{\pi}{2}-x+k2\pi\\4x-\frac{\pi}{6}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{15}+\frac{k2\pi}{5}\\x=\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)
b/
\(\Leftrightarrow cosx-\sqrt{3}sinx=sin2x-\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=sin\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{5\pi}{6}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow cos3x-\sqrt{3}sin3x=\sqrt{3}cos2x-sin2x\)
\(\Leftrightarrow\frac{1}{2}cos3x-\frac{\sqrt{3}}{2}sin3x=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(3x+\frac{\pi}{3}\right)=cos\left(2x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\3x+\frac{\pi}{3}=-2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
Giải các phương trình sau:
a, \(\sqrt{2}\) sin \(\left(2x+\frac{\pi}{4}\right)\)=3sinx+cosx+2
b, 1+sinx+cosx+sin2x+cos2x=0
c, (2cosx-1)(2sinx+cosx)=sin2x-sinx
d, cos3x+cos2x-cosx-1=0
a.
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(sinx+cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=-1\\2cosx-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)
\(\Leftrightarrow sinx\left(2cosx+1\right)+cosx\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)-sinx\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx-sinx\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx+cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
1) Giai phuong trinh:
cos2x + cosx + 1 = 0
Lời giải:
$\cos 2x+\cos x+1=0$
$\Leftrightarrow 2\cos ^2x-1+\cos x+1=0$
$\Leftrightarrow 2\cos ^2x+\cos x=0$
$\Leftrightarrow \cos x(2\cos x+1)=0$
$\Leftrightarrow \cos x=0$ hoặc $\cos x=-\frac{1}{2}$
Nếu $\cos x=0$
$\Rightarrow x=\frac{\pi}{2}+k\pi$ với $k$ nguyên.
Nếu $\cos x=-\frac{1}{2}$
$\Leftrightarrow x=\frac{2}{3}\pi +2k\pi$ hoặc $x=-\frac{2}{3}\pi +2k\pi$ với $k$ nguyên bất kỳ.
cosx + cos2x + cos3x + 1 = 0
Lời giải:
\(\cos x+\cos 2x+\cos 3x+1=0\)
\(\Leftrightarrow (\cos 2x+1)+\cos x+\cos 3x=0\)
\(\Leftrightarrow (2\cos ^2x-1+1)+2.\cos \frac{3x+x}{2}\cos \frac{3x-x}{2}=0\)
\(\Leftrightarrow 2\cos ^2x+2\cos 2x\cos x=0\)
\(\Leftrightarrow \cos x(\cos x+\cos 2x)=0\)
TH1: $\cos x=0$
$\Rightarrow x=\frac{\pi}{2}+k\pi$ (với $k$ nguyên)
TH2: $\cos x+\cos 2x=0$
$\Leftrightarrow \cos x+2\cos ^2x-1=0$
$\Leftrightarrow (2\cos x-1)(\cos x+1)=0$
$\Rightarrow \cos x=\frac{1}{2}$ hoặc $\cos x=-1$
$\Rightarrow x=\pm \frac{\pi}{3}+2k\pi$ với $k$ nguyên
Hoặc $x=(2k+1)\pi$ với $k$ nguyên