Câu 37: \(cos^22x=\sin^2\left(x+\frac{\pi}{3}\right)\)
=>\(\frac{1+cos\left(2\cdot2x\right)}{2}=\frac{1-cos\left(2x+\frac{2\pi}{3}\right)}{2}\)
=>\(cos4x=-cos\left(2x+\frac{2\pi}{3}\right)\)
=>\(cos4x=cos\left(\pi+2x+\frac{2\pi}{3}\right)=cos\left(2x+\frac53\pi\right)\)
=>\(\left[\begin{array}{l}4x=2x+\frac53\pi+k2\pi\\ 4x=-2x-\frac53\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac53\pi+k2\pi\\ 6x=-\frac53\pi+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac56\pi+k\pi\\ x=-\frac{5}{18}\pi+\frac{k\pi}{3}\end{array}\right.\)
\(x\in\left(0;\pi\right)\)
=>\(\left[\begin{array}{l}\frac56\pi+k\pi\in\left(0;\pi\right)\\ -\frac{5}{18}\pi+\frac{k\pi}{3}\in\left(0;\pi\right)\end{array}\right.\Rightarrow\left[\begin{array}{l}k+\frac56\in\left(0;1\right)\\ \frac{k}{3}-\frac{5}{18}\in\left(0;1\right)\end{array}\right.\)
=>\(\left[\begin{array}{l}k\in\left(-\frac56;\frac16\right)\\ \frac{k}{3}\in\left(\frac{5}{18};\frac{23}{18}\right)\end{array}\right.\Rightarrow\left[\begin{array}{l}k\in\left(-\frac56;\frac16\right)\\ k\in\left(\frac56;\frac{23}{6}\right)\end{array}\right.\Rightarrow\left[\begin{array}{l}k=0\\ k\in\left\lbrace1;2;3\right\rbrace\end{array}\right.\)
Khi k=0 thì \(x=\frac56\pi+k\pi=\frac56\pi+0\cdot\pi=\frac56\pi\)
Khi k=1 thì \(x=-\frac{5}{18}\pi+\frac{k\pi}{3}=-\frac{5}{18}\pi+\frac{\pi}{3}=-\frac{5}{18}\pi+\frac{6\pi}{18}=\frac{1}{18}\pi\)
Khi k=2 thì \(x=-\frac{5}{18}\pi+\frac{k\pi}{3}=-\frac{5}{18}\pi+\frac{2\pi}{3}=-\frac{5}{18}\pi+\frac{12\pi}{18}=\frac{7\pi}{18}\)
Khi k=3 thì \(x=-\frac{5}{18}\pi+\frac{k\pi}{3}=-\frac{5}{18}\pi+\pi=\frac{13}{18}\pi\)
Câu 38: \(\sin\left(x+\frac{\pi}{3}\right)+\sin5x=0\)
=>\(\sin\left(x+\frac{\pi}{3}\right)=-\sin5x=\sin\left(-5x\right)\)
=>\(\left[\begin{array}{l}-5x=x+\frac{\pi}{3}+k2\pi\\ -5x=\pi-x-\frac{\pi}{3}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}-6x=\frac{\pi}{3}+k2\pi\\ -4x=\frac23\pi+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}x=-\frac{\pi}{18}-\frac{k\pi}{3}\\ x=-\frac16\pi-\frac{k\pi}{2}\end{array}\right.\)
TH1: \(x=-\frac{\pi}{18}-\frac{k\pi}{3}\)
x\(\in\left(0;\pi\right)\)
=>\(-\frac{\pi}{18}-\frac{k\pi}{3}\in\left(0;\pi\right)\)
=>\(-\frac{k}{3}-\frac{1}{18}\in\left(0;1\right)\)
=>\(-\frac{k}{3}\in\left(\frac{1}{18};\frac{19}{18}\right)\)
=>\(k\in\left(-\frac16;-\frac{19}{6}\right)\)
=>k∈{-1;-2;-3}(1)
TH2: \(x=-\frac16\pi-\frac{k\pi}{2}\)
x\(\in\left(0;\pi\right)\)
=>\(-\frac16\pi-\frac{k\pi}{2}\in\left(0;\pi\right)\)
=>\(-\frac16-\frac{k}{2}\in\left(0;1\right)\)
=>\(-\frac{k}{2}\in\left(\frac16;\frac76\right)\)
=>\(k\in\left(-\frac13;-\frac73\right)\)
=>k∈{-1;-2}(2)
Từ (1),(2) suy ra phương trình có 5 nghiệm thuộc khoảng \(\left(0;\pi\right)\)
