Question

4sin^2+3Sinx*cosx-cosx^2x=0

Answer 1

4sin^2x+3Sinx*cosx-cosx^2x=0   делим всё на cosx^2x≠0

4tg^2x+3tgx-1=0

Замена tgx=t, t∈(-∞;+∞)

4t^2+3t-1=0

D=9+16=25

t=(-3+5)/8 = 1/4

t=(-3-5)/8 = -1

 

tgx = 1/4⇒ x=acrtg(1/4)+pik, k∈Z

tgx = -1⇒ x= -pi/4+pik, k∈Z

 

ОТВЕТ:

acrtg(1/4)+pik, k∈Z

-pi/4+pik, k∈Z