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\(\cos5x=-\sin4x\)
<=> \(\cos5x=\cos\left(4x+\frac{\pi}{2}\right)\)
\(\Leftrightarrow\orbr{\begin{cases}5x=4x+\frac{\pi}{2}+k2\pi\\5x=-4x-\frac{\pi}{2}+k2\pi\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{\pi}{2}+k2\pi\\x=-\frac{\pi}{18}+\frac{k2\pi}{9}\end{cases}}\)
Nghiệm âm lớn nhất: \(-\frac{\pi}{18}\)
Nghiệm dương nhỏ nhất: \(\frac{\pi}{2}\)
pt <=> \(\sin\left(5x+\frac{\pi}{3}\right)=\sin\left(2x-\frac{\pi}{3}+\frac{\pi}{2}\right)\)
<=> \(\sin\left(5x+\frac{\pi}{3}\right)=\sin\left(2x+\frac{\pi}{6}\right)\)
<=> \(\orbr{\begin{cases}5x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\5x+\frac{\pi}{3}=\pi-2x-\frac{\pi}{6}+k2\pi\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{\pi}{18}+\frac{k2\pi}{3}\\x=\frac{\pi}{14}+\frac{k2\pi}{7}\end{cases}}\)
Trên \(\left[0,\pi\right]\)có các nghiệm:
\(\frac{11\pi}{18},\frac{\pi}{14},\frac{5\pi}{14},\frac{9\pi}{14},\frac{13\pi}{14}\)
tính tổng:...
7.
Đặt \(\left|sinx+cosx\right|=\left|\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\right|=t\Rightarrow0\le t\le\sqrt{2}\)
Ta có: \(t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\frac{t^2-1}{2}\) (1)
Pt trở thành:
\(\frac{t^2-1}{2}+t=1\)
\(\Leftrightarrow t^2+2t-3=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
Thay vào (1) \(\Rightarrow2sinx.cosx=t^2-1=0\)
\(\Leftrightarrow sin2x=0\Rightarrow x=\frac{k\pi}{2}\)
\(\Rightarrow x=\left\{\frac{\pi}{2};\pi;\frac{3\pi}{2}\right\}\Rightarrow\sum x=3\pi\)
6.
\(\Leftrightarrow\left(1-sin2x\right)+sinx-cosx=0\)
\(\Leftrightarrow\left(sin^2x+cos^2x-2sinx.cosx\right)+sinx-cosx=0\)
\(\Leftrightarrow\left(sinx-cosx\right)^2+sinx-cosx=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx-cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\sinx-cosx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=k\pi\\x-\frac{\pi}{4}=-\frac{\pi}{4}+k\pi\\x-\frac{\pi}{4}=\frac{5\pi}{4}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=k\pi\\x=\frac{3\pi}{2}+k\pi\end{matrix}\right.\)
Pt có 3 nghiệm trên đoạn đã cho: \(x=\left\{\frac{\pi}{4};0;\frac{\pi}{2}\right\}\)
Với \(x\in\left(-\frac{\pi}{4};\frac{\pi}{2}\right)\Rightarrow cosx>0\Rightarrow3cosx+1>0\)
Do đó pt tương đương:
\(2cos2x-1=0\Rightarrow cos2x=\frac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
Pt có 2 nghiệm thuộc khoảng đã cho là \(x=\left\{-\frac{\pi}{6};\frac{\pi}{6}\right\}\)
Ta có: \(2\cos 3x(2\cos 2x + 1) = 1\)
=>\(4\cos3x\cos2x+2\cos3x=1\)
=>\(2(\cos 5x + \cos x) + 2\cos 3x = 1\)
=>\(2\cos5x+2\cos3x+2\cos x=1\) (1)
TH1: sin x=0
=>\(x=k\pi\)
Thay vào (1), ta được: \(2\cdot cos\left(5k\pi\right)+2\cdot cos\left(3k\pi\right)+2\cdot cos\left(k\pi\right)=1\)
=>2*\(\pm1\) (2*1+1)=1
=>\(\pm6\)=1(vô lý)
=>Loại
TH2: sin x<>0
(1)=>\(2\sin x \cdot \cos 5x + 2\sin x \cdot \cos 3x + 2\sin x \cdot \cos x = \sin x\)
=>\((\sin6x-\sin4x)+(\sin4x-\sin2x)+\sin2x=\sin x\)
=>sin 6x=sin x
=>\(\left[\begin{array}{l}6x=x+k2\pi\\ 6x=\pi-x+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=k2\pi\\ 7x=\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{k2\pi}{5}\\ x=\frac{\pi}{7}+\frac{k2\pi}{7}\end{array}\right.\)
TH1: \(x=\frac{k2\pi}{5}\)
\(x\in\left\lbrack-4\pi;6\pi\right\rbrack\)
=>\(-4\pi\le\frac{2k\pi}{5}\le6\pi\)
=>-10<=k<=15
Ta sẽ loại các giá trị k thỏa mãn \(x=k\pi\)
=>k∉{-10;-5;0;5;10;15}
\(S_1 = \sum_{k=-10}^{15} \frac{2k\pi}{5} - \sum_{k \in \{-10,-5,0,5,10,15\}} \frac{2k\pi}{5}\)
\(=\frac{2\pi}{5}\cdot\frac{(-10 + 15) \cdot26}{2}-\frac{2\pi}{5}\cdot15=20\pi\)
TH2: \(x=\frac{\pi+k2\pi}{7}\)
\(x \in [-4\pi; 6\pi]\)
=>\(-4\pi\le\frac{(2k+1)\pi}{7}\le6\pi\)
=>\(-29\le2k+1\le42\)
=>-15<=k<=20
Ta sẽ loại các giá trị k khiến cho x=kΩ
=>k∉{-11;-4;3;10;17}
\(S_2 = \sum_{k=-15}^{20} \frac{(2k+1)\pi}{7} - \sum_{k \in \{-11,-4,3,10,17\}} \frac{(2k+1)\pi}{7}\)
=>\(S_2 = \frac{216\pi}{7} - \frac{35\pi}{7} = \frac{181\pi}{7}\)
S=S1+S2
\(=20\pi+\frac{181\pi}{7}=\frac{321\pi}{7}\)