2.2
ĐKXĐ \(\left\{{}\begin{matrix}cos\left(\dfrac{\pi}{3}-x\right)\ne0\\cos\left(2x+\dfrac{\pi}{3}\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{\pi}{6}+k\pi\\x\ne\dfrac{\pi}{12}+\dfrac{k\pi}{2}\end{matrix}\right.\)
Phương trình:
\(\Rightarrow tan\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{tan\left(\dfrac{\pi}{3}-x\right)}\)
\(\Leftrightarrow tan\left(2x+\dfrac{\pi}{3}\right)=cot\left(\dfrac{\pi}{3}-x\right)\)
\(\Leftrightarrow tan\left(2x+\dfrac{\pi}{3}\right)=tan\left(x+\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow2x+\dfrac{\pi}{3}=x+\dfrac{\pi}{6}+k\pi\)
\(\Leftrightarrow x=-\dfrac{\pi}{6}+k\pi\) (ko thỏa mãn ĐKXĐ)
Vậy pt đã cho vô nghiệm
1.
ĐKXĐ: \(cosx-1\ne0\)
\(\Leftrightarrow x\ne k2\pi\)
b. \(2sinx-1\ne0\)
\(\Leftrightarrow sinx\ne\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{6}+k2\pi\\x\ne\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
2.1
\(\Leftrightarrow1-cos^22x-\left(1+cos2x\right)+\dfrac{3}{4}=0\)
\(\Leftrightarrow-cos^22x-cos2x+\dfrac{3}{4}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\cos2x=-\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\)
2.3
\(sinx+sin2x=cosx+cos2x\)
\(\Leftrightarrow sinx-cosx=cos2x-sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\sqrt{2}cos\left(2x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{4}\right)=sin\left(x-\dfrac{\pi}{4}\right)=cos\left(x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{4}=x+\dfrac{\pi}{4}+k2\pi\\2x+\dfrac{\pi}{4}=-x-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
2.4
\(\sqrt{3}cosx-sinx=0\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx=0\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=0\)
\(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)
2.5
\(\Leftrightarrow sinx-\sqrt{3}cosx=2\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=1\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=1\)
\(\Leftrightarrow x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\)
2.6
\(2cos2x+1=0\)
\(\Leftrightarrow cos2x=-\dfrac{1}{2}\)
\(\Leftrightarrow2x=\pm\dfrac{2\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{3}+k\pi\)
2.7
\(2sinx+cosx-2sinx.cosx-1=0\)
\(\Leftrightarrow2sinx\left(1-cosx\right)-\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(1-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=k2\pi\end{matrix}\right.\)
2.8
ĐKXĐ: \(cosx\ne-\dfrac{\sqrt{3}}{2}\) \(\Leftrightarrow x\ne\pm\dfrac{5\pi}{6}+k2\pi\)
Pt trở thành:
\(sin2x+2sin^2x-5sinx-cosx+2=0\)
\(\Leftrightarrow2sinx.cosx-cosx+\left(2sin^2x-5sinx+2\right)=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+\left(2sinx-1\right)\left(sinx-2\right)=0\)
\(\Leftrightarrow\left(sinx+cosx-2\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+2\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=-\sqrt{2}< -1\left(loại\right)\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{\pi}{6}+k2\pi\)
2.9
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\left(\sqrt{3}-1\right)cosx+\left(\sqrt{3}+1\right)sinx=4\sqrt{2}sinx.cosx\)
\(\Leftrightarrow\left(\dfrac{\sqrt{3}-1}{2\sqrt{2}}\right)cosx+\left(\dfrac{\sqrt{3}+1}{2\sqrt{2}}\right)sinx=sin2x\)
\(\Leftrightarrow cosx.sin\left(\dfrac{\pi}{12}\right)+sinx.cos\left(\dfrac{\pi}{12}\right)=sin2x\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{12}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=x+\dfrac{\pi}{12}+k2\pi\\2x=\dfrac{11\pi}{12}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k2\pi\\x=\dfrac{11\pi}{36}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
2.10
ĐKXĐ: \(\left[{}\begin{matrix}x\ne\dfrac{\pi}{2}+k2\pi\\x\ne-\dfrac{5\pi}{6}+k2\pi\\x\ne\dfrac{11\pi}{6}+k2\pi\end{matrix}\right.\)
Pt \(\Rightarrow\)\(\dfrac{cosx-sin2x}{cos2x+sinx}=\sqrt{3}\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}cos2x+\sqrt{3}sinx\)
\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos\left(2x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{5}=x+\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{6}=-x-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\left(loại\right)\\x=-\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
3a.
\(\Leftrightarrow1+cos2x+\left(m+1\right)sin2x=2m-3\)
\(\Leftrightarrow cos2x+\left(m+1\right)sin2x=2m-4\)
Pt đã cho có nghiệm khi:
\(1^2+\left(m+1\right)^2\ge\left(2m-4\right)^2\)
\(\Leftrightarrow3m^2-18m+14\le0\)
\(\Rightarrow\dfrac{9-\sqrt{39}}{3}\le m\le\dfrac{9+\sqrt{39}}{3}\)
Câu 3b là câu 51 phần trắc nghiệm hồi nãy
