`~rai~`
\(\text{Bài 1:}\\a)\sin5x+\sin3x=0\\\Leftrightarrow \sin5x=-\sin3x\\\Leftrightarrow \sin5x=\sin(-3x)\\\Leftrightarrow \left[\begin{array}{I}5x=-3x+k2\pi\\5x=\pi+3x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}8x=k2\pi\\2x=\pi+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=k\dfrac{\pi}{4}\\x=\dfrac{\pi}{2}+k\pi\end{array}\right.\\\Leftrightarrow x=k\dfrac{\pi}{4}.(k\in\mathbb{Z})\\b)\cos2x+\cos\left(3x+\dfrac{\pi}{2}\right)=0\\\Leftrightarrow 2\cos\dfrac{2x+3x+\dfrac{\pi}{3}}{2}\cos\dfrac{2x-3x-\dfrac{\pi}{3}}{2}=0\\\Leftrightarrow 2\cos\left(\dfrac{5x}{2}+\dfrac{\pi}{6}\right)\cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)=0\\\Leftrightarrow \left[\begin{array}{I}\dfrac{5x}{2}+\dfrac{\pi}{6}=0\\\dfrac{x}{2}+\dfrac{\pi}{6}=0\end{array}\right.\\\Leftrightarrow\left[\begin{array}{I}\dfrac{5x}{2}+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\\\dfrac{x}{2}+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}\dfrac{5x}{2}=\dfrac{\pi}{3}+k\pi\\\dfrac{x}{2}=\dfrac{\pi}{3}+k\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{2\pi}{15}+k\dfrac{2\pi}{5}\\x=\dfrac{2\pi}{3}+k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Bài 2:}\\a)\sin2x-1=0\\\Leftrightarrow \sin2x=1\\\Leftrightarrow 2x=\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow x=\dfrac{\pi}{4}+k\pi.(k\in\mathbb{Z})\\\text{Vì }-\pi<x<\pi\\\Leftrightarrow -\pi<\dfrac{\pi}{4}+k\pi<\pi\\\Leftrightarrow -\dfrac{5\pi}{4}<k\pi<\dfrac{3\pi}{4}\\\Leftrightarrow -\dfrac{5}{4}<k<\dfrac{3}{4}\\\text{mà k}\in\mathbb{Z}\Rightarrow k\in\{-1;0\}\\\Rightarrow x=-\dfrac{3\pi}{4};x=\dfrac{\pi}{4}.\\\text{Vậy phương trình có 2 nghiệm trên khoảng }(-\pi;\pi)\text{ là:}\\x=-\dfrac{3\pi}{4};x=\dfrac{\pi}{4}.\)
