Question

2sin^2x-5sinx*cosx=cos^2x-2

Answer 1

Ответ:

x={π/4+kπ; arctg(1/4)+kπ}, k∈Z

Объяснение:

2sin²x-5sinxcosx=cos²x-2

2sin²x-5sinxcosx=cos²x-2(cos²x+sin²x)

2sin²x-5sinxcosx=cos²x-2cos²x-2sin²x

4sin²x-5sinxcosx+cos²x=0

(4sin²x-5sinxcosx+cos²x)/cos²x=0

4tg²x-5tgx+1=0

tgx=y

4y²-5y+1=0

4y²-4y-y+1=0

4y(y-1)-(y-1)=0

(y-1)(4y-1)=0

1) y-1=0

y=1

tgx=1

x=π/4+kπ, k∈Z

2) 4y-1=0

y=1/4

tgx=1/4

x=arctg(1/4)+kπ, k∈Z