Question

sinx+cosx=0 (решите уравнение) Товарищи,в бой!))))))

Answer 1

sinx+cosx=0⇒sinx=-cosx; cosx≠0⇒
делим обе части на cosx:
tgx=-1⇒x=-π/4+πn

Answer 2

1)2sinx/2cosx/2+cos²x/2-sin²x/2=0/cos²x/2≠0
tg²x/2-2tgx/2-1=0
tgx/2=a
a²-2a-1=0
D=4+4=8
a1=(2-2√2)/2=1-√2⇒tgx/2=1-√2⇒x/2=arctg(1-√2)+πn⇒x=2arctg(1-√2)+2πn
a2=1+√2⇒tgx/2=1+√2⇒x/2=arctg(1+√2)+πn⇒x=2arctg(1+√2)+2πn

2)√2(√2/2sinx+√2/2cosx)=0
√2sin(x+π/4)=0
sin(x+π/4)=0
x+π/4=πn
x=-π/4+πn
выбирай