Question

16cos^4x -24cos^2x +9=0 (2pi; 3pi)

Answer 1

16cos⁴x-24cos²x+9=0       x∈(2π;3π)
cos²x=v  ⇒
16v²-24v+9=0
(4v-3)²=0
4v-3=0
v=3/4
cos²x=3/4
cosx=+/-√(3/4)=+/-√3/2
x₁=π/6+2πn  x₂=-π/6+2πn.
Так как х∈(2π;3π) ⇒
х=π/6+2πn.