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a.
\(cos\left(3x-\frac{\pi}{6}\right)=sin\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(3x-\frac{\pi}{6}\right)=cos\left(\frac{\pi}{6}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=\frac{\pi}{6}-2x+k2\pi\\3x-\frac{\pi}{6}=2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\cos3x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne\frac{1}{2}\end{matrix}\right.\)
\(tan3x-tanx=0\)
\(\Leftrightarrow\frac{sin3x}{cos3x}-\frac{sinx}{cosx}=0\)
\(\Leftrightarrow sin3x.cosx-cos3x.sinx=0\)
\(\Leftrightarrow sin2x=0\)
\(\Leftrightarrow2sinx.cosx=0\)
\(\Leftrightarrow sinx=0\Leftrightarrow x=k\pi\)
c.
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{2\pi}{5}\right)=\frac{1}{2}-\frac{1}{2}cos\left(4x+\frac{8\pi}{5}\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=-cos\left(4x+\frac{3\pi}{5}+\pi\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=cos\left(4x+\frac{3\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{3\pi}{5}=2x-\frac{2\pi}{5}+k2\pi\\4x+\frac{3\pi}{5}=\frac{2\pi}{5}-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
d.
\(\Leftrightarrow cos^2\left(2x-1\right)=0\)
\(\Leftrightarrow cos\left(2x-1\right)=0\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{1}{2}+\frac{k\pi}{2}\)
1d.
Đề ko rõ
1e.
\(\Leftrightarrow\left(4cos^3x-3cosx\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(4cos^2x-3\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(2cos2x-1\right)^2cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left[\left(2cos2x-1\right)^2.cos2x-1\right]=0\)
\(\Leftrightarrow cos^2x\left(4cos^32x-4cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow cos^2x\left(cos2x-1\right)\left(4cos^22x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)
2b.
Đề thiếu
2c.
Nhận thấy \(cos2x=0\) ko phải nghiệm, chia 2 vế cho \(cos^32x\)
\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)
\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)
\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)
\(\Leftrightarrow...\)
a/ Thiếu đề, sau dấu "-" hình như còn gì đó
b/ \(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=sin\left(\frac{\pi}{4}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
c/ \(\Rightarrow sin2x=-sinx\Leftrightarrow sin2x=sin\left(-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=-x+k2\pi\\2x=\pi+x+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{3}\\x=\pi+k2\pi\end{matrix}\right.\)
d/ \(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2\left(sinx.cosx\right)^2=1\)
\(\Leftrightarrow sinx.cosx=0\Leftrightarrow sin2x=0\)
\(\Rightarrow2x=k\pi\Rightarrow x=\frac{k\pi}{2}\)
e/ f/ Thiếu đề
g/ \(\Leftrightarrow\left[{}\begin{matrix}cos3x=cos2x\\cos3x=-cos2x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cos3x=cos2x\\cos3x=cos\left(\pi-2x\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=2x+k2\pi\\3x=-2x+k2\pi\\3x=\pi-2x+k2\pi\\3x=2x-\pi+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{k2\pi}{5}\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\\x=-\pi+k2\pi\end{matrix}\right.\)
e/
ĐKXĐ: ...
\(\Leftrightarrow\frac{1}{cos^2x}\left(9-13cosx\right)+4=0\)
\(\Leftrightarrow\frac{9}{cos^2x}-\frac{13}{cosx}+4=0\)
Đặt \(\frac{1}{cosx}=t\)
\(\Rightarrow9t^2-13t+4=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=\frac{4}{9}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{1}{cosx}=1\\\frac{1}{cosx}=\frac{4}{9}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{9}{4}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=k2\pi\)
d/
\(\Leftrightarrow cos^22x+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{\pi}{2}\right)-1=0\)
\(\Leftrightarrow1-sin^22x+\frac{1}{2}sin2x-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^22x+sin2x+1=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\\2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
Câu 2 bạn coi lại đề
3.
\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)
\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)
\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
4.
Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm
5.
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)
\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
1.
\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Leftrightarrow...\\sinx+cosx+sinx.cosx-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
2.
\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)
Xét (1):
Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm
10. ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
\(2cos2x+tanx=\frac{4}{5}\)
\(\Leftrightarrow4cos^2x-2+tanx=\frac{4}{5}\)
\(\Leftrightarrow\frac{4}{1+tan^2x}+tanx-\frac{14}{5}=0\)
Đặt \(tanx=t\)
\(\Rightarrow\frac{20}{1+t^2}+5t-14=0\)
\(\Leftrightarrow5t^3-14t^2+5t+6=0\)
\(\Leftrightarrow\left(t-2\right)\left(5t^2-4t-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{2+\sqrt{19}}{5}\\t=\frac{2-\sqrt{19}}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tanx=2=tana\\tanx=\frac{2+\sqrt{19}}{5}=tanb\\tanx=\frac{2-\sqrt{19}}{5}=tanc\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=a+k\pi\\x=b+k\pi\\x=c+k\pi\end{matrix}\right.\)
9.
\(\Leftrightarrow cos2x-3cosx=2\left(cosx+1\right)\)
\(\Leftrightarrow2cos^2x-1-3cosx=2cosx+2\)
\(\Leftrightarrow2cos^2x-5cosx-3=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=3\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)
8.
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\frac{1}{cos^2x}-1+3cot^2x-4=0\)
\(\Leftrightarrow\frac{sin^2x}{cos^2x}+3cot^2x-4=0\)
\(\Leftrightarrow tan^2x+3cot^2x-4=0\)
Đặt \(cot^2x=t\ge0\)
\(\Rightarrow\frac{1}{t}+3t-4=0\Leftrightarrow3t^2-4t+1=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cotx=\pm1\\cotx=\pm\frac{1}{\sqrt{3}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{4}+k\pi\\x=\pm\frac{\pi}{3}+k\pi\end{matrix}\right.\)
6. ĐKXĐ \(cosx\ne0\)
\(9-13cosx+\frac{4}{1+tan^2x}=0\)
\(\Leftrightarrow9-13cosx+4cos^2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{9}{4}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=k2\pi\)
7. ĐKXĐ \(sinx\ne0\)
\(\frac{1}{sin^2x}=cotx+3\)
\(\Leftrightarrow\frac{1}{sin^2x}-1-cotx-2=0\)
\(\Leftrightarrow\frac{cos^2x}{sin^2x}-cotx-2=0\)
\(\Leftrightarrow cot^2x-cotx-2=0\Rightarrow\left[{}\begin{matrix}cotx=-1\\cotx=2=cota\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=a+k\pi\end{matrix}\right.\)
5. ĐKXĐ: ...
\(\Leftrightarrow\frac{3}{cosx}+tan^2x+1-10=0\)
\(\Leftrightarrow\frac{1}{cos^2x}+\frac{3}{cosx}-10=0\)
Đặt \(\frac{1}{cosx}=a\Rightarrow\left|a\right|\ge1\)
\(a^2+3a-10=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{cosx}=2\\\frac{1}{cosx}=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\cosx=-\frac{1}{5}=cos\alpha\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=\pm\alpha+k2\pi\end{matrix}\right.\)
3.
\(\Leftrightarrow4cos^5\left(2-6x\right)+8\left[cos\left(2-6x\right)+1\right]-13=0\)
Đặt \(cos\left(2-6x\right)=a\Rightarrow\left|a\right|\le1\)
\(\Rightarrow4a^5+8a-5=0\)
Bạn coi lại đề bài, pt bậc 5 ko nhẩm được nghiệm thì làm sao mà giải?
4.
ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow1+tan^2x-\left(3+\sqrt{3}\right)tanx-3+\sqrt{3}=0\)
\(\Leftrightarrow tan^2x-\left(3+\sqrt{3}\right)tanx-2+\sqrt{3}=0\)
\(\Delta=\left(3+\sqrt{3}\right)^2-4\left(-2+\sqrt{3}\right)=20+2\sqrt{3}\)
Chắc bạn lại nhầm hệ số nào đó, kết quả xấu kinh dị
\(\Rightarrow\left[{}\begin{matrix}tanx=\frac{3+\sqrt{3}-\sqrt{20+2\sqrt{3}}}{2}=tana\\tanx=\frac{3+\sqrt{3}+\sqrt{20+2\sqrt{3}}}{2}=tanb\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=a+k\pi\\x=b+k\pi\end{matrix}\right.\)
2.
\(\Leftrightarrow2cos^2x-1+9cosx+5=0\)
\(\Leftrightarrow2cos^2x+9cosx+4=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow cosx=cos\left(\frac{2\pi}{3}\right)\)
\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)
1.
\(4\left(1-cos^23x\right)+2\left(\sqrt{3}+1\right)cos3x-\sqrt{3}-4=0\)
Đặt \(cos3x=a\Rightarrow\left|a\right|\le1\)
\(\Rightarrow4\left(1-a^2\right)+2\left(\sqrt{3}+1\right)a-\sqrt{3}-4=0\)
\(\Leftrightarrow-4a^2+2\left(\sqrt{3}+1\right)a-\sqrt{3}=0\)
\(\Delta'=\left(\sqrt{3}+1\right)^2-4\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}a=\frac{-\sqrt{3}-1+\sqrt{3}-1}{-4}=\frac{1}{2}\\a=\frac{-\sqrt{3}-1-\sqrt{3}+1}{-4}=\frac{\sqrt{3}}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cos3x=\frac{1}{2}\\cos3x=\frac{\sqrt{3}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=\pm\frac{\pi}{3}+k2\pi\\3x=\pm\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{9}+\frac{k2\pi}{3}\\x=\pm\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)