Question

помогите решить уравнение.

Answer 1

4sin²x+8sin (3π/2 + x ) + 1 =0
4sin²x+8(-cosx)+1=0
4sin²x-8cosx+1=0
4sin²x-8cosx=-1
4(1cos²x)-8cosx+1=0
4-4cos²x-8cosx+1=0
5-4cos²x-8cosx=0

5-4t²-8t=0
-4t²-8t+5=0
4t²-8t-5=0
8²-4*4*(-5) =64+80= 144 =12
t = - 8 + 12 / 8 = 1/2
t= -8 - 12 / 8 = - 5 /2

cosx=1/2
x=π/3 + 2kπ , k∈Z
X=5π/3 + 2kπ , k∈Z

cosx=-5/2
x∉R

тогда ответ:
x = π/3 + 2kπ , k∈Z
5π/3 + 2kπ , k∈Z