Question

25^cos(3п/2+x)=5^1-cos2x
Корни на отрезке: [-5п; -3п/2]

Answer 1

5^2cos(3π/2+x)=5^1-cos2x
2cos(3π/2+x)=1-cos2x
2sinx=2sin²x
2sin²x-2sinx=0
2sinx(sinx-1)=0
sinx=0⇒x=πn,n∈z
-5π≤πn≤-3π/2
-5≤n≤-3/2
n=-5⇒x=-5π
n=-4⇒x=-4π
n=-3⇒x=-3π
n=-2⇒x=-2π
sinx=1⇒x=π/2+2πk,k∈z
-5π≤π/2+2πk≤-3π/2
-10≤1+4k≤-2
-11≤4k≤-3
-11/4≤k≤-3/4
k=-2⇒x=π/2-4π=--7π/2
n=-1⇒x=π/2-2π=-3π/2