Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ \(m=0\) pt vô nghiêm
Với \(m\ne0\Rightarrow cosx=\frac{m+1}{m}\)
\(-1\le cosx\le1\Rightarrow-1\le\frac{m+1}{m}\le1\)
\(\Rightarrow m\le-\frac{1}{2}\)
b/ \(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)-cos4x=m\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x-cos4x=m\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x-\left(1-2sin^22x\right)=m\)
\(\Leftrightarrow\frac{5}{4}sin^22x=m\)
Do \(0\le\frac{5}{4}sin^22x\le\frac{5}{4}\Rightarrow0\le m\le\frac{5}{4}\)
c/ \(\Leftrightarrow1-\frac{3}{4}sin^22x=m\left(1-\frac{1}{4}sin^22x\right)\)
\(\Leftrightarrow\left(m-3\right)sin^22x=4m-4\)
- Với \(m=3\) pt vô nghiệm
- Với \(m\ne3\Rightarrow sin^22x=\frac{4m-4}{m-3}\)
Do \(0\le sin^22x\le1\Rightarrow0\le\frac{4m-4}{m-3}\le1\)
\(\Rightarrow\frac{1}{3}\le m\le1\)
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
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.
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow1-2sin^22x=0\)
\(\Leftrightarrow cos4x=0\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)
a/ \(\Leftrightarrow2cosx.cos2x=cos2x\)
\(\Leftrightarrow2cosx.cos2x-cos2x=0\)
\(\Leftrightarrow cos2x\left(2cosx-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cos2x=0\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
b/ \(\Leftrightarrow2sinx.sin2x=sinx\)
\(\Leftrightarrow2sinx.sin2x-sinx=0\)
\(\Leftrightarrow sinx\left(2sin2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=0\\sin2x=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\2x=\frac{\pi}{6}+k2\pi\\2x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)
c/ \(\Leftrightarrow sin3x-sinx+sin4x-sin2x=0\)
\(\Leftrightarrow2cos2x.sinx+2cos3x.sinx=0\)
\(\Leftrightarrow sinx\left(cos2x+cos3x\right)=0\)
\(\Leftrightarrow2sinx.2cos\frac{5x}{2}.cos\frac{x}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=0\\cos\frac{5x}{2}=0\\cos\frac{x}{2}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k2\pi\\\frac{x}{2}=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{5}+\frac{k4\pi}{5}\\x=\pi+k4\pi\end{matrix}\right.\)
d/ \(\Leftrightarrow sin3x-sinx-\left(sin4x-sin2x\right)=0\)
\(\Leftrightarrow2cos2x.sinx-2cos3x.sinx=0\)
\(\Leftrightarrow sinx\left(cos2x-cos3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos2x=cos3x\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\2x=3x+k2\pi\\2x=-3x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{k2\pi}{5}\end{matrix}\right.\)
1: \(\Leftrightarrow\sin^3x=-\cos^3x\)
\(\Leftrightarrow\sin^3x=-\sin^3\left(\dfrac{\Pi}{2}-x\right)\)
\(\Leftrightarrow\sin^3x=\sin^3\left(-\dfrac{\Pi}{2}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\Pi}{2}+x+k2\Pi\\x=\dfrac{\Pi}{2}-x+k2\Pi\end{matrix}\right.\Leftrightarrow x=\dfrac{\Pi}{4}+k\Pi\)
2: \(\Leftrightarrow-\dfrac{1}{2}\sin x+\dfrac{\sqrt{3}}{2}\cos x=0\)
\(\Leftrightarrow\sin x\cdot\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\cdot\cos x=0\)
\(\Leftrightarrow\sin x\cdot\dfrac{\cos\Pi}{6}-\cos x\cdot\sin\left(\dfrac{\Pi}{6}\right)=0\)
\(\Leftrightarrow\sin\left(x-\dfrac{\Pi}{6}\right)=0\)
\(\Leftrightarrow x-\dfrac{\Pi}{6}=k\Pi\)
hay \(x=k\Pi+\dfrac{\Pi}{6}\)
\(sina+sinb+sinc+3=0\)
\(\Leftrightarrow\left(sina+1\right)+\left(sinb+1\right)+\left(sinc+1\right)=0\)
Do \(\left\{{}\begin{matrix}sina\ge-1\\sinb\ge-1\\sinc\ge-1\end{matrix}\right.\) ;\(\forall a;b;c\)
\(\Rightarrow\left(sina+1\right)+\left(sinb+1\right)+\left(sinc+1\right)\ge0\)
Dấu "=" xảy ra khi và chỉ khi \(sina=sinb=sinc=-1\)
\(\Rightarrow cosa=cosb=cosc=0\Rightarrow cosa+cosb+cosc+10=10\)
b/ \(sinx=1-sin^2x\Rightarrow sinx=cos^2x\)
\(\Rightarrow sin^2x=cos^4x\Rightarrow1-cos^2x=cos^4x\)
\(\Rightarrow cos^4x+cos^2x=1\Rightarrow\left(cos^4x+cos^2x\right)^2=1\)
\(\Rightarrow cos^8x+2cos^6x+cos^4x=1\)