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Lời giải:
PT $\Leftrightarrow (2\cos x-1)(2\sin x+\cos x)=2\sin x\cos x-\sin x$
$\Leftrightarrow (2\cos x-1)(2\sin x+\cos x)=\sin x(2\cos x-1)$
$\Leftrightarrow (2\cos x-1)(\sin x+\cos x)=0$
$\Rightarrow 2\cos x=1$ hoặc $\sin x=-\cos x=\cos (\pi -x)=\sin (x-\frac{\pi}{2})$
Đến đây thì đơn giản rồi.
Nkjuiopmli Sv5: Bạn chuyển vế sin x(2cos x-1) sang vế trái thì vế phải còn 0 đó.
2.1
a.
\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
b.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
1.
a, Phương trình có nghiệm khi:
\(\left(m+2\right)^2+m^2\ge4\)
\(\Leftrightarrow m^2+4m+4+m^2\ge4\)
\(\Leftrightarrow2m^2+4m\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}m\ge0\\m\le-2\end{matrix}\right.\)
b, Phương trình có nghiệm khi:
\(m^2+\left(m-1\right)^2\ge\left(2m+1\right)^2\)
\(\Leftrightarrow2m^2+6m\le0\)
\(\Leftrightarrow-3\le m\le0\)
2.
a, Phương trình vô nghiệm khi:
\(\left(2m-1\right)^2+\left(m-1\right)^2< \left(m-3\right)^2\)
\(\Leftrightarrow4m^2-4m+1+m^2-2m+1< m^2-6m+9\)
\(\Leftrightarrow4m^2-7< 0\)
\(\Leftrightarrow-\dfrac{\sqrt{7}}{2}< m< \dfrac{\sqrt{7}}{2}\)
b, \(2sinx+cosx=m\left(sinx-2cosx+3\right)\)
\(\Leftrightarrow\left(m-2\right)sinx-\left(2m+1\right)cosx=-3m\)
Phương trình vô nghiệm khi:
\(\left(m-2\right)^2+\left(2m+1\right)^2< 9m^2\)
\(\Leftrightarrow m^2-4m+4+4m^2+4m+1< 9m^2\)
\(\Leftrightarrow m^2-1>0\)
\(\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\)
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.
\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)
\(\Leftrightarrow sinx\left(2cosx+1\right)+cosx\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
a.\(\dfrac{sin2x+cosx-\sqrt{3}\left(cos2x+sinx\right)}{2sin2x-\sqrt{3}}=1\left(1\right)\)
ĐKXĐ: sin2x≠\(\dfrac{\sqrt{3}}{2}\)
(1) ⇔ sin2x + cosx - \(\sqrt{3}\) ( cos2x + sinx) = 2sin2x - \(\sqrt{3}\)
⇔cosx - \(\sqrt{3}\) sinx = \(\sqrt{3}\) cos2x + sin2x +\(\sqrt{3}\)
⇔\(\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}\)
⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=sin\left(2x+\dfrac{\Pi}{3}\right)-sin\dfrac{\Pi}{3}\)
⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=2cos\left(x+\dfrac{\Pi}{3}\right)sinx\)
⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=2sin\left(\dfrac{\Pi}{6}-x\right)sinx\)
⇔\(sin\left(\dfrac{\Pi}{6}-x\right)\left(2sinx-1\right)=0\)
Đến đây tự giải tiếp nha nhớ đối chiếu đk.
b.\(\left(2cosx-1\right)cotx=\dfrac{3}{sinx}+\dfrac{2sinx}{cosx-1}\left(1\right)\)
ĐKXĐ: sinx≠0 và cosx≠1
(1)⇔\(\left(2cosx-1\right)\dfrac{cosx}{sinx}=\dfrac{3}{sinx}+\dfrac{2sinx}{cosx-1}\)
⇔cosx(2cosx-1)(cosx-1) = 3(cosx-1) + 2sin2x
⇔2cos3x - cos2x - 2cosx +1 = 0
⇔ (cosx-1)(cosx+1)(2cosx-1)=0