4sin2x-3sin(2x-pi/2)=5

Question

4sin2x-3sin(2x-pi/2)=5

Answer 1

4sin2x-3sin(2x-π/2)=5
4sin2x+3cos2x=5
8sinxcosx + 3cos²x-3sin²x=5sin²x+5cos²x
8sin²x-8sinxcosx+2cos²x=0|:2cos²x
4tg²x-4tgx+1=0
(2tgx-1)²=0
tgx=1/2
x=arctg(1/2) + πn,n ∈ Z

Answer 2

4sin2x+3cos2x=5
Разделим обе части уравнения на √(4²+3²)=√(16+9)=√25=5
4/5*sin2x+3/5*cos2x=1
sin(2x+φ)=1
2x+φ=π/2+2πn,n∈Z
2x=π/2-φ+2πn,n∈Z
x=π/4-1/2*φ+πn,n∈Z
φ=arcsin3/5
x=π/4-1/2*arcsin0,6+πn,n∈Z