Ta có: \(\sin5x+\sqrt3\cdot cos5x=2\cdot\sin7x\)
=>\(\frac12\cdot\sin5x+\frac{\sqrt3}{2}\cdot cos5x=\sin7x\)
=>\(\sin\left(5x+\frac{\pi}{3}\right)=\sin7x\)
=>\(\left[\begin{array}{l}7x=5x+\frac{\pi}{3}+k2\pi\\ 7x=\pi-5x-\frac{\pi}{3}+k2\pi=-5x+\frac23\pi+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}2x=\frac{\pi}{3}+k2\pi\\ 12x=\frac23\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+k\pi\\ x=\frac{1}{18}\pi+\frac{k\pi}{6}\end{array}\right.\)
TH1: \(x=\frac{\pi}{6}+k\pi\)
\(x\in\left(0;\frac{\pi}{2}\right)\)
=>\(\frac{\pi}{6}+k\pi\in\left(0;\frac{\pi}{2}\right)\)
=>\(k+\frac16\in\left(0;\frac12\right)\)
=>k∈(-1/6;1/3)
mà k là số nguyên
nên k=0(1)
TH2: \(x=\frac{\pi}{18}+\frac{k\pi}{6}\)
\(x\in\left(0;\frac{\pi}{2}\right)\)
=>\(\frac{\pi}{18}+\frac{k\pi}{6}\in\left(0;\frac{\pi}{2}\right)\)
=>\(\frac{k}{6}+\frac{1}{18}\in\left(0;\frac12\right)\)
=>k/6∈(-1/18;8/18)
=>k∈(-6/18;48/18)
=>k∈(-1/3;8/3)
mà k là số nguyên
nên k∈{0;1;2}(2)
Từ (1),(2) suy ra phương trình có 1+3=4 nghiệm trên khoảng (0;pi/2)
giúp mk với ạ
