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a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp
b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)
\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)
\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)
\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)
c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:
\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)
Đặt \(\sqrt{tanx+1}=t\ge0\)
\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)
\(\Leftrightarrow3t^3-5t^2+3t-10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)
d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)\)
Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
Chứng minh các biểu thức đã cho không phụ thuộc vào x.
Từ đó suy ra f'(x)=0
a) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
b) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
c) f(x)=\(\frac{1}{4}\)(\(\sqrt{2}\)-\(\sqrt{6}\))=>f'(x)=0
d,f(x)=\(\frac{3}{2}\)=>f'(x)=0
c/
\(\Leftrightarrow1-sin^22x+\sqrt{3}sin2x+sin2x=1+\sqrt{3}\)
\(\Leftrightarrow-sin^22x+\left(\sqrt{3}+1\right)sin2x-\sqrt{3}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=\sqrt{3}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
d/
\(\Leftrightarrow4\left(1-2sin^2x\right)+5sinx=4\left(3sinx-4sin^3x\right)+5\)
\(\Leftrightarrow16sin^3x-8sin^2x-7sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(4sinx+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=arcsin\left(-\frac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\frac{1}{4}\right)+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow3cos^2x+4sin\left(2\pi-\frac{\pi}{2}-x\right)+1=0\)
\(\Leftrightarrow3cos^2x-4sin\left(x+\frac{\pi}{2}\right)+1=0\)
\(\Leftrightarrow3cos^2x-4cosx+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm arcos\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)
a: \(\sin3x+cos2x=1+2\cdot\sin x\cdot cos2x\)
=>sin3x+cos2x=1+sin(x+2x)+sin(x-2x)
=>sin3x+cos2x=1+sin3x-sin x
=>cos2x-1+sin x=0
=>\(1-2\cdot\sin^2x-1+\sin x=0\)
=>\(-2\cdot\sin^2x+\sin x=0\)
=>sin x(2sin x-1)=0
TH1: sin x=0
=>\(x=k\pi\)
TH2: 2sin x-1=0
=>\(\sin x=\frac12\)
=>\(\left[\begin{array}{l}x=\frac{\pi}{6}+k2\pi\\ x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\)
b: \(\sin^3x+cos^3x=2\cdot\left(\sin^5x+cos^5x\right)\)
=>\(\sin^3x-2\cdot\sin^5x+cos^3x-2\cdot cos^5x=0\)
=>\(\sin^3x\left(1-2\cdot\sin^2x\right)+cos^3x\left(1-2\cdot cos^2x\right)=0\)
=>\(\sin^3x\cdot cos2x-cos^3x\cdot cos2x=0\)
=>\(cos2x\left(\sin^3x-cos^3x\right)=0\)
TH1: cos2x=0
=>\(2x=\frac{\pi}{2}+k\pi\)
=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)
TH2: \(\sin^3x-cos^3x=0\)
=>\(\sin^3x=cos^3x\)
=>sin x=cosx
=>\(\sin x-cosx=0\)
=>\(\sqrt2\cdot\sin\left(x-\frac{\pi}{4}\right)=0\)
=>\(\sin\left(x-\frac{\pi}{4}\right)=0\)
=>\(x-\frac{\pi}{4}=k\pi\)
=>\(x=\frac{\pi}{4}+k\pi\)
f: ĐKXĐ: \(\begin{cases}\sin x<>0\\ cosx<>0\end{cases}\Rightarrow\begin{cases}x<>k\pi\\ x<>\frac{\pi}{2}+k\pi\end{cases}\Rightarrow x<>\frac{k\pi}{2}\)
\(\frac{\tan x-\sin x}{\sin^3x}=\frac{1}{cosx}\)
=>\(\frac{\frac{\sin x}{cosx}-\sin x}{\sin^3x}=\frac{1}{cosx}\)
=>\(\frac{\frac{1}{cosx}-1}{\sin^2x}=\frac{1}{cosx}\)
=>\(\sin^2x=cosx\cdot\left(\frac{1}{cosx}-1\right)=1-cosx\)
=>\(1-cos^2x=1-cosx\)
=>\(cos^2x-cosx=0\)
=>cosx(cosx-1)=0
TH1: cosx=0
=>\(x=\frac{\pi}{2}+k\pi\) (loại)
TH2: cosx-1=0
=>cosx=1
=>\(x=k2\pi\)
=>sin x=0
=>Loại