Question

1.$sin^{2} -6sinx*cosx=0$
2.$tgx-5ctgx=4$
3.$2cos^{2}*4x-cos4x\geq 0$

Answer 1

Ответ:

Объяснение:

1. sin²x-6sinxcosx=0

sinx(sinx-6cosx)=0

sinx=0   sinx=6cosx

x=Пk;     sinx/cosx=6

             tgx=6  x=arctg6+Пk

2.tgx-5ctgx=4

tgx-5/tgx-4=0

tg²x-4tgx-5=0

tgx=2+-√(4+5)=2+-3

tgx=5

x=arctg5+Пk

tgx=-1

x=-П/4+Пk

3.

2cos²4x-cos4x>=0

cos4x(2cos4x-1)>=0

cos4x<=0    cos4x>=1/2

4x [П/2+Пk;3П/2+Пk]

4x [-П/3+2пk;П/3+2Пk]

x∈ [П/8+Пk/4;3П/8+Пk/4] U  [-П/12+Пk.2;П/12+Пk/2]