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e/
\(2cos^2x+2cos^22x+4cos^32x-3cos2x=5\)
\(\Leftrightarrow1+cos2x+2cos^22x+4cos^32x-3cos2x=5\)
\(\Leftrightarrow2cos^32x+cos^22x-cos2x-2=0\)
\(\Leftrightarrow\left(cos2x-1\right)\left(2cos^22x+3cos2x+2\right)=0\)
\(\Leftrightarrow cos2x=1\)
\(\Leftrightarrow x=k\pi\)
\(d\text{) }4\left(sin^4x+cos^4x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow4\left(1-2sin^2x\cdot cos^2x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow-8sin^2x\cdot cos^2x+\sqrt{3}sin4x=-2\\ \Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\\ \Leftrightarrow cos4x-1+\sqrt{3}sin4x=-2\\ \Leftrightarrow\frac{1}{2}cos4x+\frac{\sqrt{3}}{2}sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\frac{\pi}{6}\cdot cos4x+cos\frac{\pi}{6}\cdot sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=sin\frac{-\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=\frac{-\pi}{6}+a2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{12}+\frac{a\pi}{2}\\x=\frac{\pi}{4}+\frac{b\pi}{2}\end{matrix}\right.\)
\(e\text{) }4sinx\cdot cosx\cdot cos2x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}cos4x=1\\ \Leftrightarrow sin4x\cdot cos\frac{\pi}{4}+cos4x\cdot sin\frac{\pi}{4}=1\\ \Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=1=sin\frac{\pi}{2}\\ \Leftrightarrow4x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{\pi}{16}+\frac{k\pi}{2}\)
\(\text{a) }cos^2x+sin2x-1=0\\ \Leftrightarrow2sinx\cdot cosx-sin^2x=0\\ \Leftrightarrow sinx\left(2cosx-sinx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=2cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=a\pi\\x=arctan\left(2\right)+b\pi\end{matrix}\right.\)
\(\text{b) }\sqrt{3}sin2x+cos^4x-sin^4x=\sqrt{2}\\ \Leftrightarrow\sqrt{3}sin2x+\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=\sqrt{2}\\ \Leftrightarrow\frac{\sqrt{3}}{2}\cdot sin2x+\frac{1}{2}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\frac{\pi}{4}\\ \\ \Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=\frac{\pi}{4}+a2\pi\\2x+\frac{\pi}{6}=\frac{3\pi}{4}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}+a\pi\\x=\frac{7\pi}{24}+b\pi\end{matrix}\right.\)
\(c\text{) }cos^2x-sin^2x=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\\ \Leftrightarrow cos^2x-sin^2x=\sqrt{2}\left(sinx\cdot\frac{\sqrt{2}}{2}+cosx\cdot\frac{\sqrt{2}}{2}\right)\\ \Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)=sinx+cosx\\ \Leftrightarrow\left[{}\begin{matrix}cosx-sinx=1\\sinx=-cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos^2x+\left(cosx-1\right)^2=1\\tanx=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\\tanx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+a\pi\\x=b2\pi\\x=\frac{3\pi}{4}=c\pi\end{matrix}\right.\)
d/
ĐKXĐ: ...
\(\Leftrightarrow cos^2x+\frac{1}{cos^2x}+2=2\left(cosx+\frac{1}{cosx}\right)\)
\(\Leftrightarrow\left(cosx+\frac{1}{cosx}\right)^2=2\left(cox+\frac{1}{cosx}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx+\frac{1}{cosx}=0\\cosx+\frac{1}{cosx}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos^2x+1=0\left(vn\right)\\cos^2x-2cosx+1=0\end{matrix}\right.\)
\(\Rightarrow cosx=1\)
\(\Rightarrow x=k2\pi\)
c/
\(\Leftrightarrow cos\frac{6x}{5}+2=3cos\frac{4x}{5}\)
Đặt \(\frac{2x}{5}=a\)
\(\Rightarrow cos3a+2=3cos2a\)
\(\Leftrightarrow4cos^3a-3cosa+2=6cos^2a-3\)
\(\Leftrightarrow4cos^3a-6cos^2a-3cosa+5=0\)
\(\Leftrightarrow\left(cosa-1\right)\left(4cos^2a-2cosa-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosa=1\\cosa=\frac{1+\sqrt{21}}{4}>1\left(l\right)\\cosa=\frac{1-\sqrt{21}}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cos\left(\frac{2x}{5}\right)=1\\cos\left(\frac{2x}{5}\right)=\frac{1-\sqrt{21}}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2x}{5}=k2\pi\\\frac{2x}{5}=\pm arccos\left(\frac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k5\pi\\x=\pm\frac{5}{2}arccos\left(\frac{1-\sqrt{21}}{4}\right)+k5\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=1+cos\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow1-3sin^2x.cos^2x=1+sin2x\)
\(\Leftrightarrow-\frac{3}{4}sin^22x=sin2x\)
\(\Leftrightarrow3sin^22x+4sin2x=0\)
\(\Leftrightarrow sin2x\left(3sin2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sin2x=-\frac{4}{3}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\frac{k\pi}{2}\)
a/
\(\Leftrightarrow cos2x=sin3x\)
\(\Leftrightarrow cos2x=cos\left(\frac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-3x+k2\pi\\2x=3x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{10}+\frac{k2\pi}{5}\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow\left(sinx-1\right)\left(2sinx+1\right)\left(sin^2x-2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\frac{1}{2}\\sinx=1-\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow x=...\)
6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
5.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
\(cos^4x+sin^4x=1+\frac{1}{2}sin4x\)
\(\Leftrightarrow\left(cos^2x+sin^2x\right)^2-2\left(sinx.cosx\right)^2=1+\frac{1}{2}sin4x\)
\(\Leftrightarrow1-\frac{1}{2}sin^22x=1+\frac{1}{2}sin4x\)
\(\Leftrightarrow sin4x+sin^22x=0\)
\(\Leftrightarrow2sin2x.cos2x+sin^22x=0\)
\(\Leftrightarrow sin2x\left(2cos2x+sin2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=0\Rightarrow x=\frac{k\pi}{2}\\2cos2x+sin2x=0\left(1\right)\end{matrix}\right.\)
Xét (1)
\(\Leftrightarrow\frac{1}{\sqrt{5}}sin2x+\frac{2}{\sqrt{5}}cos2x=0\)
Đặt \(cosa=\frac{1}{\sqrt{5}}\) với \(a\in\left[0;\pi\right]\)
\(\Rightarrow sin2x.cosa+cos2x.sina=0\)
\(\Leftrightarrow sin\left(2x+a\right)=0\)
\(\Rightarrow2x+a=k\pi\Rightarrow x=-\frac{a}{2}+\frac{k\pi}{2}\)
\(2\left(cos^2x-1\right)=sinx.cos3x\)
\(\Leftrightarrow-2sin^2x=sinx.cos3x\)
\(\Leftrightarrow sinx.cos3x+2sin^2x=0\)
\(\Leftrightarrow sinx\left(cos3x+2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos3x+2sinx=0\left(1\right)\end{matrix}\right.\)
Bạn có ghi nhầm đề ko nhỉ, pt (1) dù giải được nhưng khá khó đấy, phải vận dụng công thức nhân 3 và nghiệm ko hề đẹp