\(sinx+4cosx=2+sin2x\)
\(\left(1-sin2x\right)\left(sinx+cosx\right)=cos2x\)
\(1+sinx+cosx+sin2x+cos2x=0\)
\(sinx+sin2x+sin3x=1+cosx+cos2x\)
\(sin^22x-cos^28x=sin\left(\dfrac{17\pi}{2}+10x\right)\)
\(sinx+4cosx=2+sin2x\)
\(\left(1-sin2x\right)\left(sinx+cosx\right)=cos2x\)
\(1+sinx+cosx+sin2x+cos2x=0\)
\(sinx+sin2x+sin3x=1+cosx+cos2x\)
\(sin^22x-cos^28x=sin\left(\dfrac{17\pi}{2}+10x\right)\)
Cho hàm số \(y=\sin2x+\cos2x+3.\) GTLN của hàm số trên\(\left[-\dfrac{\pi}{4};\dfrac{\pi}{4}\right]\) là số \(a+b\sqrt{2}.\) . Tính \(a+b\)
\(y=\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)+3\)
Do \(sin\left(2x+\dfrac{\pi}{4}\right)\le1\Rightarrow y\le3+\sqrt{2}\)
\(\Rightarrow a=3;b=1\Rightarrow a+b=\)
24. Tìm GTLN của hàm số: \(y=3\cos\left(x-\dfrac{\pi}{2}\right)+1\)
26. a) Tìm GTLN của hàm số: \(y=\cos2x+\sin2x\)
b) Giải PT: \(\sin x+\sqrt{3}\cos x=1\)
24.
\(cos\left(x-\dfrac{\pi}{2}\right)\le1\Rightarrow y\le3.1+1=4\)
\(y_{max}=4\)
26.
\(y=\sqrt{2}cos\left(2x-\dfrac{\pi}{4}\right)\)
Do \(cos\left(2x-\dfrac{\pi}{4}\right)\le1\Rightarrow y\le\sqrt{2}\)
\(y_{max}=\sqrt{2}\)
b.
\(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
Giải PT
a1) \(3.\cos4x-2^{ }\cos^23x=1\)
a2) \(2\cos2x-8\cos x+7=\dfrac{1}{\cos x}\)
a3) \(\dfrac{\left(1+\sin x+\cos2x\right)\sin\left(x+\dfrac{\pi}{4}\right)}{1+\tan x}=\dfrac{1}{\sqrt{2}}\cos x\)
a4) \(9\sin x+6\cos x-3\sin2x+\cos2x=8\)
a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)
\(\Leftrightarrow3cos4x-cos6x-2=0\)
Đặt \(t=2x\)
Pttt:\(3cos2t-cos3t-2=0\)
\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)
\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))
\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)
\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)
\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))
Vậy...
a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)
Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)
\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)
\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))
\(\Rightarrow sinx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)
Vậy...
Giải các phương trình sau:
a, sinx+cosx+1+sin2x+cos2x=0
b, sinx(1+cos2x)+sin2x=1+cos2x
c, \(\frac{1}{sinx}+\frac{1}{sin\left(x-\frac{3\pi}{2}\right)}=4sin\left(\frac{7\pi}{4}-x\right)\)
d, sin4x+cos4x=\(\frac{7}{8}cot\left(x+\frac{\pi}{3}\right)cot\left(\frac{\pi}{6}-x\right)\)
@Nguyễn Việt Lâm giúp em với ạ
a.
\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
b.
\(sinx\left(1+2cos^2x-1\right)+2sinx.cosx=1+2cos^2x-1\)
\(\Leftrightarrow cos^2x.sinx+sinx.cosx-cos^2x=0\)
\(\Leftrightarrow cosx\left(sinx.cosx+sinx-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\sinx.cosx+sinx-cosx=0\left(1\right)\end{matrix}\right.\)
Xét (1), đặt \(sinx-cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{1-t^2}{2}\end{matrix}\right.\)
\(\Rightarrow\frac{1-t^2}{2}+t=0\)
\(\Leftrightarrow-t^2+2t+1=0\Rightarrow\left[{}\begin{matrix}t=1-\sqrt{2}\\t=1+\sqrt{2}>\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1-\sqrt{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
giải các pt
a) \(sin2x-2\sqrt{3}cos^2x=4cosx\)
b) \(sin^2x-3cos^2x=sinx-\sqrt{3}cosx\)
c) \(sin6x\left(cos3x-1\right)-sin6x.sin3x=0\)
d) \(\left(sin2x-cos2x\right)^2-3\left(sin2x-cos2x\right)-4=0\)
e) \(\frac{sin2x+sin6x}{cos2x}-2cos4x=2\sqrt{2}\)
a/
\(\Leftrightarrow2sinx.cosx-2\sqrt{3}cos^2x-4cosx=0\)
\(\Leftrightarrow2cosx\left(sinx-\sqrt{3}cosx-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\sinx-\sqrt{3}cosx=2\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=1\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=1\)
\(\Leftrightarrow x-\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{5\pi}{6}+k2\pi\)
b/
\(\Leftrightarrow\left(sinx-\sqrt{3}cosx\right)\left(sinx+\sqrt{3}cosx\right)=sinx-\sqrt{3}cosx\)
\(\Leftrightarrow\left(sinx-\sqrt{3}cosx\right)\left(sinx+\sqrt{3}cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\sqrt{3}cosx\left(1\right)\\sinx+\sqrt{3}cosx=1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow tanx=\sqrt{3}\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
\(\left(2\right)\Leftrightarrow\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow sin6x\left(cos3x-1-sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin6x=0\Rightarrow x=\frac{k\pi}{6}\\cos3x-sin3x=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow sin3x-cos3x=-1\)
\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\3x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{3}\\x=\frac{\pi}{2}+\frac{k2\pi}{3}\end{matrix}\right.\)
GPT: \(\dfrac{\left(\sin x-\cos x\right)\left(\sin2x-3\right)-\sin2x-\cos2x+1}{2\sin x-\sqrt{2}}=0\)
ĐKXĐ: \(sinx\ne\dfrac{\sqrt{2}}{2}\)
\(\left(sinx-cosx\right)\left(sin2x-3\right)+\left(sinx-cosx\right)^2+\left(sin^2x-cos^2x\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sin2x-3\right)+\left(sinx-cosx\right)^2+\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sin2x-3+2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\\left(sin2x-1\right)+2\left(sinx+1\right)=0\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=-\dfrac{\pi}{4}+k2\pi\)
tìm tập xác định của hàm số
a) y = \(\dfrac{25}{cos\left(3x+\dfrac{\pi}{4}\right)}\)
b) y = \(\dfrac{cos2x-4}{sin3x+1}\)
c) y = \(\dfrac{4+cos5x}{sin2x-1}\)
1, Giải phương trình :
a, sin2x - 2cos2x = 0
b, \(sin\left(4x+\frac{1}{2}\right)=\frac{1}{3}\)
c, \(sin^4x+cos^4x=\frac{3}{4}\)
d,\(\left(cosx-sinx\right)^2=1-cos3x\)
e,\(\left(cosx+sinx\right)^2=3sin2x\)
2. Phương trình : \(sin3x=cos^4x-sin^4x\) có tập nghiệm trùng với tập nghiệm cua phương trình nào sau đây :
A. cos2x = sin3x B. cos2x = -sin3x C. cos2x = sin2x D. cos2x = -sin2x
a.\(\frac{k\Pi}{2}+\frac{\alpha}{2}\)
b.\(\left\{{}\begin{matrix}x=\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\\x=\Pi-\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\end{matrix}\right.\)
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 \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)