Question

решить: sin2x-3cos4x=4

Answer 1

sin2x-3cos4x=4

sin2x-3(1-2sin²2x)=4

6sin²2x+sin2x-7=0

sin2x=a, -1≤a≤1 

6a²+a-7=0

a₁=1, a₂=-7/6

sin2x=1

2x=-π/2 +2πk

x= -π/4 + πk