Question

2cos 2x-5sin^2 x+7=0

Answer 1

2cos2x - 5sin²x+7=0
cos2x=1-2sin²x
2*(1-2sin²x) - 5sin²x+7=0
2-4sin²x-5sin²x+7=0
-9sin²x=-9
sin²x=1
sinx=+-1 
x=+-π/2 + 2πk k∈Z ;( или π/2 + πk ,k∈Z )