8cos^2x + 10sinx - 1 = 0

подробнее

8 (1-sin²x) + 10sinx-1=0

8-8sin²x+10sinx-1=0

-8sin²x+10sinx+7=0

8sin²x-10sinx-7=0

пусть sinx=t. t∈[-1.1]

8t²-10t-7=0 D=10²-4*8*-7=100+224=324

t1=10-18/16=-8/16=-1/2 t2=10+18/16=28/16=7/4 - посторонний корень

sinx=-1/2 x = (-1) ^k arcsin (-1/2) + πk. k∈Z

x = (-1) ^ (k+1) π/6+πk. k∈Z