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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/
\(\Leftrightarrow cos^22x+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{\pi}{2}\right)-1=0\)
\(\Leftrightarrow1-sin^22x+\frac{1}{2}sin2x-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^22x+sin2x+1=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\\2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x-2\left(1-sin^22x\right)=0\)
\(\Leftrightarrow1-\frac{1}{2}\left(cos6x+cos2x\right)-2cos^22x=0\)
\(\Leftrightarrow1-cos4x.cos2x-2cos^22x=0\)
\(\Leftrightarrow2cos^22x-1+cos4x.cos2x=0\)
\(\Leftrightarrow cos4x+cos4x.cos2x=0\)
\(\Leftrightarrow cos4x\left(cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
d/
ĐKXĐ: \(sin2x\ne0\) \(\Leftrightarrow2x\ne k\pi\)
\(\Leftrightarrow1+\frac{cos2x}{sin2x}=\frac{1-cos2x}{sin^22x}\)
\(\Leftrightarrow sin^22x+sin2x.cos2x=1-cos2x\)
\(\Leftrightarrow sin^22x-1+sin2x.cos2x+cos2x=0\)
\(\Leftrightarrow-cos^22x+sin2x.cos2x+cos2x=0\)
\(\Leftrightarrow cos2x\left(sin2x-cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin2x-cos2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=k\pi\left(l\right)\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)
e/
\(\Leftrightarrow1+cos2x+1+cos4x+1+cos6x=3+3cosx.cos4x\)
\(\Leftrightarrow cos2x+cos6x+cos4x-3cosx.cos4x=0\)
\(\Leftrightarrow2cos4x.cos2x+cos4x-3cosx.cos4x=0\)
\(\Leftrightarrow cos4x\left(2cos2x+1-3cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\Rightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\\2cos2x-3cosx+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\left(2cos^2x-1\right)-3cosx+1=0\)
\(\Leftrightarrow4cos^2x-3cosx-1=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm arccos\left(-\frac{1}{4}\right)+k2\pi\end{matrix}\right.\)
d/
\(\Leftrightarrow5\left(1+cosx\right)=2+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow5\left(1+cosx\right)=2+sin^2x-cos^2x\)
\(\Leftrightarrow5+5cosx=2+1-cos^2x-cos^2x\)
\(\Leftrightarrow2cos^2x+5cosx+2=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)
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.
\(\Leftrightarrow\left(1-cos6x\right)cos2x+1-cos2x=0\)
\(\Leftrightarrow cos2x-cos2x.cos6x+1-cos2x=0\)
\(\Leftrightarrow\frac{1}{2}\left(cos8x-cos4x\right)-1=0\)
\(\Leftrightarrow2cos^24x-cos4x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-1\\cos4x=\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow4x=\pi+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
3.
Đặt \(\frac{x}{6}=t\Rightarrow\frac{1}{4}+cos^22t=\frac{1}{2}sin^23t\)
\(\Leftrightarrow1+4cos^22t=1-cos6t\)
\(\Leftrightarrow cos6t+4cos^22t=0\)
\(\Leftrightarrow4cos^32t+4cos^22t-3cos2t=0\)
\(\Leftrightarrow cos2t\left(4cos^22t+4cos2t-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2t=0\\cos2t=\frac{1}{2}\\cos2t=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{\pi}{4}+\frac{k\pi}{2}\\t=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{3}=\frac{\pi}{4}+\frac{k\pi}{2}\\\frac{x}{3}=\frac{\pi}{6}+k\pi\\\frac{x}{3}=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
e/
\(\Leftrightarrow3\left(1-cos6x\right)-\left(2cos^26x-1\right)=4\)
\(\Leftrightarrow-2cos^26x-3cos6x=0\)
\(\Leftrightarrow cos6x\left(2cos6x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cos6x=0\\cos6x=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow6x=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{12}+\frac{k\pi}{3}\)
d/
\(\Leftrightarrow3\left(1-cos2x\right)-2\left(1-cos^22x\right)=5\)
\(\Leftrightarrow2cos^22x-3cos2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\frac{3+\sqrt{41}}{4}\left(l\right)\\cos2x=\frac{3-\sqrt{41}}{4}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{1}{2}arccos\left(\frac{3-\sqrt{41}}{4}\right)+k\pi\)
Nghiệm xấu quá :(