Question

4cos2x+3sin2x-5=0
A){arctg3}
B){arctg1/3+pn,n€Z}
C){p/4+pn, n€Z}
D){arctg3+pn,n€Z}

Answer 1

Ответ:

Объяснение:

4cos2x+3sin2x=5

-5 + 4cos(2x) + 3sin(2x) = 0

- (cosx - 3sinx)^2 = 0

(cosx - 3sinx)^2 = 0

(cosx - 3sinx) = 0

cosx = 3sinx    /cosx ≠ 0

tgx = 1/3

x = arctg(1/3) + πn, n∈Z