Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐKXĐ: ...
a/ \(\frac{sin2x}{cos2x}+\frac{cosx}{sinx}=8cos^2x\)
\(\Leftrightarrow sin2x.sinx+cos2x.cosx=8cos^2x.sinx.cos2x\)
\(\Leftrightarrow cosx=4sin2x.cos2x.cosx\)
\(\Leftrightarrow cosx=2sin4x.cosx\)
\(\Leftrightarrow cosx\left(2sin4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin4x=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)
b/ \(\frac{cosx}{sinx}-\frac{sinx}{cosx}+4sin2x=\frac{1}{sinx.cosx}\)
\(\Leftrightarrow cos^2x-sin^2x+4sin2x.sinx.cosx=1\)
\(\Leftrightarrow cos2x+2sin^22x=1\)
\(\Leftrightarrow cos2x+2\left(1-cos^22x\right)=1\)
\(\Leftrightarrow-2cos^22x+cos2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)
1c/
\(5sinx-2=3\left(1-sinx\right)\frac{sin^2x}{1-sin^2x}\)
\(\Leftrightarrow5sinx-2=\frac{3sin^2x}{1+sinx}\)
\(\Leftrightarrow\left(5sinx-2\right)\left(1+sinx\right)=3sin^2x\)
\(\Leftrightarrow5sin^2x+3sinx-2=3sin^2x\)
\(\Leftrightarrow2sin^2x+3sinx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-2\left(l\right)\end{matrix}\right.\) \(\Rightarrow x=...\)
Bài 2:
a/ \(\Leftrightarrow\frac{\left(m+1\right)\left(1-cos2x\right)}{2}-sin2x+cos2x=0\)
\(\Leftrightarrow2sin2x+\left(m-1\right)cos2x=m+1\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(4+\left(m-1\right)^2\ge\left(m+1\right)^2\)
\(\Leftrightarrow4m\le4\Rightarrow m\le1\)
Nhân 2 vế với \(sin4x\) sau đó tách:
\(\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=\frac{2sin2x.cos2x}{cosx}+\frac{2sin2x.cos2x}{sin2x}=\frac{4sinx.cosx.cos2x}{cosx}+\frac{2sin2x.cos2x}{sin2x}\)
Rồi rút gọn
3.
Theo điều kiện của pt lượng giác bậc nhất:
\(m^2+\left(3m+1\right)^2\ge\left(1-2m\right)^2\)
\(\Leftrightarrow10m^2+6m+1\ge4m^2-4m+1\)
\(\Leftrightarrow3m^2+5m\ge0\Rightarrow\left[{}\begin{matrix}m\ge0\\m\le-\frac{5}{3}\end{matrix}\right.\)
4.
\(\Leftrightarrow1-sin^2x-\left(m^2-3\right)sinx+2m^2-3=0\)
\(\Leftrightarrow-sin^2x-m^2sinx+2m^2+3sinx-2=0\)
\(\Leftrightarrow\left(-sin^2x+3sinx-2\right)+m^2\left(2-sinx\right)=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2-sinx\right)+m^2\left(2-sinx\right)=0\)
\(\Leftrightarrow\left(2-sinx\right)\left(sinx-1+m^2\right)=0\)
\(\Leftrightarrow sinx=1-m^2\)
\(\Rightarrow-1\le1-m^2\le1\)
\(\Rightarrow m^2\le2\Rightarrow-\sqrt{2}\le m\le\sqrt{2}\)
1.
Bạn xem lại đề, \(sin^2x\left(\frac{x}{2}-\frac{\pi}{4}\right)\) là sao nhỉ?Có cả x trong lẫn ngoài ngoặc?
2.
ĐKXĐ: \(sinx\ne0\)
\(\left(2sinx-cosx\right)\left(1+cosx\right)=sin^2x\)
\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=1-cos^2x\)
\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)-\left(1+cosx\right)\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1+cosx\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
a.
\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
a.
\(1-sin^2x+1-2sin^2x+sinx+2=0\)
\(\Leftrightarrow-3sin^2x+sinx+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{4}{3}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=-\frac{\pi}{2}+k2\pi\)
b. ĐKXĐ; ...
\(5tanx-\frac{2}{tanx}-3=0\)
\(\Leftrightarrow5tan^2x-3tanx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=-\frac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-\frac{2}{5}\right)+k\pi\end{matrix}\right.\)
e.
Ko rõ vế phải
f.
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow1-2sin^22x=0\)
\(\Leftrightarrow cos4x=0\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)
1: \(\Leftrightarrow4\cdot\dfrac{1+\cos2x}{2}-6\cdot\dfrac{1-\cos2x}{2}+5\sin2x-4=0\)
\(\Leftrightarrow2+2\cos2x-3+3\cos2x+5\sin2x-4=0\)
\(\Leftrightarrow5\sin2x+5\cos2x=5\)
\(\Leftrightarrow\cos2x+\sin2x=1\)
\(\Leftrightarrow\sqrt{2}\cdot\sin\left(2x+\dfrac{\Pi}{4}\right)=1\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{4}=\dfrac{\Pi}{4}+k2\Pi\\2x+\dfrac{\Pi}{4}=\dfrac{3\Pi}{4}+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
2: \(\Leftrightarrow\sqrt{3}\cdot\dfrac{1+\cos2x}{2}+\sin2x-\sqrt{3}\cdot\dfrac{1-\cos2x}{2}-1=0\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}\cos2x+\sin2x+\sqrt{3}\cdot\dfrac{\cos2x-1}{2}-1=0\)
\(\Leftrightarrow\sin2x+\dfrac{\sqrt{3}}{2}\cos2x+\dfrac{\sqrt{3}}{2}\cos2x-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}-2}{2}=0\)
\(\Leftrightarrow\sin2x+\sqrt{3}\cos2x=\dfrac{\sqrt{3}-\sqrt{3}+2}{2}=1\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{3}=\dfrac{\Pi}{6}+k2\Pi\\2x+\dfrac{\Pi}{3}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{12}\Pi+k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
a/ \(y=sin2x+\left(\sqrt{3}+1\right)cos2x+sin^2x-cos^2x-1\)
\(=sin2x+\sqrt{3}cos2x-1=2sin\left(2x+\frac{\pi}{3}\right)-1\)
Do \(-1\le sin\left(2x+\frac{\pi}{3}\right)\le1\Rightarrow-3\le y\le1\)
b/ \(y=2sin^2x-2cos^2x-3sinx.cosx-1\)
\(=-2cos2x-\frac{3}{2}sin2x-1=-\frac{5}{2}\left(\frac{3}{5}sinx+\frac{4}{5}cosx\right)-1\)
\(=-\frac{5}{2}sin\left(x+a\right)-1\Rightarrow-\frac{7}{2}\le y\le\frac{3}{2}\)
c/ \(y=1-sin2x+2cos2x+\frac{3}{2}sin2x=\frac{1}{2}sin2x+2cos2x+1\)
\(=\frac{\sqrt{17}}{2}\left(\frac{1}{\sqrt{17}}sin2x+\frac{4}{\sqrt{17}}cos2x\right)+1=\frac{\sqrt{17}}{2}sin\left(2x+a\right)+1\)
\(\Rightarrow-\frac{\sqrt{17}}{2}+1\le y\le\frac{\sqrt{17}}{2}+1\)
d.
\(\Leftrightarrow\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=0\)
\(\Leftrightarrow sin^2x-cos^2x=0\)
\(\Leftrightarrow-cos2x=0\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
e. Đề thiếu
f.
\(\Leftrightarrow sin2x=\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\left(cos^2\frac{x}{2}+sin^2\frac{x}{2}\right)\)
\(\Leftrightarrow sin2x=cos^2\frac{x}{2}-sin^2\frac{x}{2}\)
\(\Leftrightarrow sin2x=cosx\)
\(\Leftrightarrow sin2x=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-x+k2\pi\\2x=x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
a.
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\sqrt{2}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=-\frac{\pi}{2}+k2\pi\)
b.
\(\Leftrightarrow sin2x=1\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
c.
\(\Leftrightarrow2sin2x.cos2x=-1\)
\(\Leftrightarrow sin4x=-1\)
\(\Leftrightarrow4x=-\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)