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\(cos^4x-sin^4x=sin3x+cos4x\)
\(\Leftrightarrow\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=sin3x+cos4x\)
\(\Leftrightarrow cos2x=sin3x+cos4x\)
\(\Leftrightarrow cos4x-cos2x+sin3x=0\)
\(\Leftrightarrow-2sin3x.sinx+sin3x=0\)
\(\Leftrightarrow sin3x\left(1-2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=\left\{0;\dfrac{\pi}{3};\dfrac{2\pi}{3};\pi;\dfrac{\pi}{6};\dfrac{5\pi}{6}\right\}\)
\(\Rightarrow\sum x=3\pi\)
\(\Leftrightarrow2cos5x.cosx=2cos5x.sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos5x=0\\cosx=sin2x=cos\left(\frac{\pi}{2}-2x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{2}-2x+k2\pi\\x=2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{10}+\frac{k\pi}{5}\\x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
1/ ĐKXĐ: \(\cos2x\ne0\)
\(\frac{\cos4x}{\cos2x}=\frac{\sin2x}{\cos2x}\)\(\Leftrightarrow\cos4x-\sin2x=0\)
\(\Leftrightarrow2\cos^22x-1-\sin2x=0\)
\(\Leftrightarrow2-2\sin^22x-1-\sin2x=0\)
\(\Leftrightarrow2\sin^22x+\sin2x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\frac{1}{2}=\sin\frac{\pi}{6}\\\sin2x=-1=\sin\frac{-\pi}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{6}+2k\pi\\2x=\frac{5\pi}{6}+2k\pi\\2x=\frac{-\pi}{2}+2k\pi\left(l\right)\\2x=\frac{3\pi}{2}+2k\pi\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)
2/ \(\sin2.4x+\cos4x=1+2\sin2x.\cos\left(2x+4x\right)\)
\(\Leftrightarrow2\sin4x.\cos4x+\cos4x=1+2\sin2x.\left(\cos2x.\cos4x-\sin2x.\sin4x\right)\)
\(\Leftrightarrow2\sin4x.\cos4x+\cos4x=1+2\sin2x.\cos2x.\cos4x-2\sin^22x.\sin4x\)
\(\Leftrightarrow2\sin4x.\cos4x+\cos4x=1+\sin4x.\cos4x-\sin4x+\cos4x.\sin4x\)
Đến đây bn tự giải nốt nhé, lm kiểu bthg thôi bởi vì đã quy về hết sin4x và cos4x r
\(\Leftrightarrow3sinx-4sin^3x+4cos^3x-3cosx+2cosx=0\)
\(\Leftrightarrow3sinx-cosx-4sin^3x+4cos^3x=0\)
Với \(cosx=0\) ko phải nghiệm, với \(cosx\ne0\) chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow3tanx\left(1+tan^2x\right)-\left(1+tan^2x\right)-4tan^3x+4=0\)
\(\Leftrightarrow-tan^3x-tan^2x+3tanx+3=0\)
\(\Leftrightarrow-tan^2x\left(tanx+1\right)+3\left(tanx+1\right)=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(3-tan^2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)
Tới đây chắc bạn hoàn thành được phần còn lại
\(cosx+cos3x+cos2x+cos4x=0\)
\(\Leftrightarrow2cos2x.cosx+2cos3x.cosx=0\)
\(\Leftrightarrow cosx.\left(cos2x+cos3x\right)=0\)
\(\Leftrightarrow cosx.cos\frac{5x}{2}.cos\frac{x}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos\frac{5x}{2}=0\\cos\frac{x}{2}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\\x=\pi+k2\pi\end{matrix}\right.\)
\(sinx+sin7x+sin3x+sin5x=0\)
\(\Leftrightarrow2sin4x.cos3x+2sin4x.cosx=0\)
\(\Leftrightarrow sin4x\left(cos3x+cosx\right)=0\)
\(\Leftrightarrow sin4x.cos2x.cosx=0\)
\(\Leftrightarrow sin4x=0\)
\(\Rightarrow4x=k\pi\Rightarrow x=\frac{k\pi}{4}\)
Lý do chỉ cần 1 pt sin4x=0 do sin4x bao hàm cả cosx và cos2x ở trong đó
cos 6x+cos4x=sin7x-sin3x
=>2*cos5x*cosx=2*cos5x*sin2x
=>cos5x(cosx-sin2x)=0
=>cos5x=0 hoặc sin2x=sin(pi/2-x)
=>5x=pi/2+kpi hoặc 2x=pi/2-x+k2pi hoặc 2x=pi/2+x+k2pi
=>x=pi/10+kpi/5; x=pi/6+k2pi/3; x=pi/2+k2pi