Question

cos2 x + 2 sen2 x - 3/2=0

Answer 1

Cos2 x + 2 sen2 x - 3/2=0

[1-tg²(x)]/[1+tg²(x)] +[2tg(x)]/[1+tg²(x)] - 3/2=0

[1-tg²(x)+2tg(x)]/[1+tg²(x)] - 3/2=0

[1-tg²(x)+2tg(x)]/[1+tg²(x)] - 3/2=0

2 - 2tg²(x) + 4tg(x) - 3 - 3tg²x =0
5tg²x+4tgx-1=0
tgx = -1
tgx = tg(3π/4)
A tangente de dois arcos são iguais quando esses arcos são côngruos ou explementares.
côngruos --> x =3π/4+kπ

explementares --> x-3π/4 = π
x = 3π/4 + π
x = 7π/4