Question

sin2x-cos2x=0
уравнение решите(

Answer 1

$sin2x-cos2x=0 \ \ frac{sin2x}{cos2x} - frac{cos2x}{cos2x} = 0 \ \ tg2x - 1 = 0 \ \ tg2x = 1 \ \ 2x = arctg1 + pi n, n in Z \ \ 2x = frac{ pi }{4} + pi n, n in Z \ \ x = frac{ pi }{8} + frac{pi n}{2}, n in Z$

Ответе: $x = frac{ pi }{8} + frac{pi n}{2}, n in Z$