Question

Найти значения Х заключенных между 0 и pi/2

Answer 1

$\displaystyle\bf\\Sin2x=\frac{1}{2}\\\\\\\left[\begin{array}{ccc}2x=\dfrac{\pi }{6}+2\pi n ,n\in Z\\2x=\dfrac{5\pi }{6} +2\pi n,n\in Z\end{array}\right.\\\\\\\left[\begin{array}{ccc}x=\dfrac{\pi }{12}+\pi n ,n\in Z\\x=\dfrac{5\pi }{12} +\pi n,n\in Z\end{array}\right.\\\\\\1) \ \ x=\frac{\pi }{12} +\pi n\\\\\\0 < \frac{\pi }{12} +\pi n < \frac{\pi }{2} \\\\\\0 < \pi +12\pi n < 6\pi \\\\\\0 < 1+12n < 6\\\\\\-1 < 12n < 5\\\\\\-\frac{1}{12} < n < \frac{5}{12}$

$\displaystyle\bf\\n=0 \ \ \ \Rightarrow \ \ \ x=\frac{\pi }{12} \\\\\\2)\\\\x=\frac{5\pi }{12} +\pi n,n\in Z\\\\\\0 < \frac{5\pi }{12} +\pi n < \frac{\pi }{2}\\\\\\0 < 5\pi +12\pi n < 6\pi \\\\\\0 < 5+12n < 6\\\\\\-5 < 12n < 1\\\\\\-\frac{5}{12} < n < \frac{1}{12} \\\\\\n=0 \ \ \ \Rightarrow \ \ \ x=\frac{5\pi }{12} \\\\\\Otvet \ : \ x=\frac{\pi }{12} \ \ \ ; \ \ \ x=\frac{5\pi }{12}$