Question

5 cos ²x+6sin x -6=0​

Answer 1

Ответ: x∈{π/2+2πn; arcsin0.2+2πn; π-arcsin0.2+2πn} n∈Z

Объяснение:

5 cos ²x+6sin x -6=0​     5 cos ²x=5-5sin²x

=> ​5-5sin²x+6sinx-6=0

-5sin²x+6sinx-1=0

5sin²x-6sinx+1=0

sinx=t

5t²-6t+1=0

D=36-20=16

t1=(6+4)/10=1   t2(6-4)/10= 1/5=0.2

sinx=1 => x=π/2+2πn , n∈Z

sinx=0.2 => x2=arcsin0.2+2πn

x3=π-arcsin0.2+2πn

Ответ x∈{π/2+2πn; arcsin0.2+2πn; π-arcsin0.2+2πn} n∈Z