Question

Найти корни уравнения sinx+cosx=0

Answer 1

sinx+cosx=0
sinx=-cosx
cos(pi/2 - x + 2n pi)=cos(pi+x)
pi/2 - x + 2n pi = pi+x
x=-pi/4 + n pi
 interval [0;2pi]  n=1, i  2
x=(3/4)pi   i   x=(7/4)pi

Answer 2

$sinx+cosx=0/:cosx \neq 0\\x \neq \frac{ \pi }{2} + \pi n\\\\ \frac{sinx}{cosx} + \frac{cosx}{cosx} =0\\tgx+1=0\\tgx=-1\\x=- \frac{ \pi }{4} + \pi n$
n ∈ Z