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a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp
b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)
\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)
\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)
\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)
c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:
\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)
Đặt \(\sqrt{tanx+1}=t\ge0\)
\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)
\(\Leftrightarrow3t^3-5t^2+3t-10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)
d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)\)
Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
5.
\(\Leftrightarrow sin\left(2cosx\right)=1\)
\(\Leftrightarrow2cosx=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow cosx=\frac{\pi}{4}+k\pi\)
Do \(-1\le cosx\le1\Rightarrow-1\le\frac{\pi}{4}+k\pi\le1\)
Mà \(k\in Z\Rightarrow k=0\)
\(\Rightarrow cosx=\frac{\pi}{4}\)
\(\Leftrightarrow x=\pm arccos\left(\frac{\pi}{4}\right)+k2\pi\)
3.
\(\Leftrightarrow sin2x+1=2\left(\frac{1-cos2x}{2}\right)\)
\(\Leftrightarrow sin2x+cos2x=0\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow2x+\frac{\pi}{4}=k\pi\)
\(\Leftrightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)
4. ĐKXĐ; ...
\(\Leftrightarrow\frac{sinx.cos2x}{cosx.sin2x}+1=0\)
\(\Leftrightarrow sinx.cos2x+cosx.sin2x=0\)
\(\Leftrightarrow sin3x=0\)
\(\Leftrightarrow3sinx-4sin^3x=0\)
\(\Leftrightarrow3-4sin^2x=0\)
\(\Leftrightarrow3-2\left(1-cos2x\right)=0\)
\(\Leftrightarrow cos2x=-\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow1-sin^22x+\sqrt{3}sin2x+sin2x=1+\sqrt{3}\)
\(\Leftrightarrow-sin^22x+\left(\sqrt{3}+1\right)sin2x-\sqrt{3}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=\sqrt{3}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
d/
\(\Leftrightarrow4\left(1-2sin^2x\right)+5sinx=4\left(3sinx-4sin^3x\right)+5\)
\(\Leftrightarrow16sin^3x-8sin^2x-7sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(4sinx+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=arcsin\left(-\frac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\frac{1}{4}\right)+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow3cos^2x+4sin\left(2\pi-\frac{\pi}{2}-x\right)+1=0\)
\(\Leftrightarrow3cos^2x-4sin\left(x+\frac{\pi}{2}\right)+1=0\)
\(\Leftrightarrow3cos^2x-4cosx+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm arcos\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)
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...
1.
ĐKXĐ: \(x\ne k\pi\)
\(\Leftrightarrow\left(2cos2x-1\right)\left(sinx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\sinx=3>1\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
2. Bạn kiểm tra lại đề, pt này về cơ bản ko giải được.
3.
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\dfrac{3\left(sinx+\dfrac{sinx}{cosx}\right)}{\dfrac{sinx}{cosx}-sinx}-2cosx=2\)
\(\Leftrightarrow\dfrac{3\left(1+cosx\right)}{1-cosx}+2\left(1+cosx\right)=0\)
\(\Leftrightarrow\left(1+cosx\right)\left(\dfrac{3}{1-cosx}+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(loại\right)\\cosx=\dfrac{5}{2}\left(loại\right)\end{matrix}\right.\)
Vậy pt đã cho vô nghiệm
