Question

Решить тригонометрический пример

Answer 1

$\displaystyle\bf\\Sin2x=2Cosx\cdot Sin5x\\\\Sin2x-2Cosx\cdot Sin5x=0\\\\2Sinx Cosx-2Cosx\cdot Sin5x=0\\\\2Cosx\cdot(Sinx-Sin5x)=0\\\\2Cosx\cdot 2Sin\frac{x-5x}{2} Cos\frac{x+5x}{2} =0\\\\Cosx\cdot Sin2x\cdot Cos3x=0\\\\\\\left[\begin{array}{ccc}Cosx=0\\Sin2x=0\\Cos3x=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\dfrac{\pi }{2}+\pi n,n\in Z \\x=\pi n ,n\in Z\\3x=\dfrac{\pi }{2}+\pi n,n\in Z\end{array}\right$

$\displaystyle\bf\\\left[\begin{array}{ccc}x=\dfrac{\pi }{2}+\pi n,n\in Z \\x=\dfrac{\pi n}{2} ,n\in Z\\x=\dfrac{\pi }{6}+\dfrac{\pi n}{3} ,n\in Z\end{array}\right\\\\\\1)\\\\x=\dfrac{\pi }{2}+\pi n\\\\4\pi \leq \dfrac{\pi }{2}+\pi n\leq 5\pi \ |:\pi \\\\4\leq \frac{1}{2}+n\leq 5\\\\3,5\leq n\leq 4,5\\\\n=4 \ \ \ \Rightarrow \ \ x=\frac{\pi }{2} +4\pi =\boxed{\frac{9\pi }{2}} \\\\2) \\\\x=\frac{\pi n}{2} \\\\4\pi \leq \dfrac{\pi n }{2}\leq 5\pi \\\\4\leq \frac{n}{2}\leq 5$

$\displaystyle\bf\\8\leq n\leq 10\\\\n=8 \ \ \ \Rightarrow \ \ \ x=\boxed{4\pi} \\\\n=9 \ \ \ \Rightarrow \ \ \ x=\boxed{\frac{9\pi }{2}} \\\\n=10 \ \ \ \Rightarrow \ \ \ x=\boxed{5\pi }\\\\3)\\\\x=\frac{\pi }{6}+\frac{\pi n}{3} \\\\4\pi \leq \frac{\pi }{6}+\frac{\pi n}{3} \leq 5\pi \\\\24\pi \leq \pi +2\pi n\leq 30\pi \\\\24\leq 1+2n\leq 30\\\\23\leq 2n\leq 29\\\\11,5\leq n\leq 14,5$

$\displaystyle\bf\\n=12 \ \ \ \Rightarrow \ \ \ x=\frac{\pi }{6}+\frac{\pi \cdot 12}{3}=\frac{\pi }{6}+4\pi=\boxed{\frac{25\pi }{6} }\\\\n=13 \ \ \ \Rightarrow \ \ \ x=\frac{\pi }{6}+\frac{\pi \cdot 13}{3}=\boxed{\frac{27\pi }{6} }\\\\n=14 \ \ \ \Rightarrow \ \ \ x=\frac{\pi }{6}+\frac{\pi \cdot 14}{3}=\boxed{\frac{29\pi }{6}}$