Question

Log3(cosx+sin2x+9)=2

Answer 1

Ответ: π/2 + πk, - π/6 + 2πk, - 5π/6 + 2πk.

Answer 2

Cosx+sin2x+9=3^2
Cosx+sin2x+9=9
Cosx+sin2x=0
Cosx+cosx*sinx=0
Cosx(1+sinx)=0
1+sinx=0
Sinx=-1
X=270градусов
Sin270=-1