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

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

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