Question

Помогите решить,срочно

Answer 1

1
2sin²x+sin²2x=2
2(1-cos2x)/2+(1-cos4x)/2=2
2-2cos2x+1-cos4x=4
4-3+2cos2x+2cos²2x-1=0
2cos²2x+2cos2x=0
2cos2x(cos2x+1)=0
cos2x=0⇒2x=π/2+πk⇒x=π/4+πk/2
-π/2≤π/4+πk/2≤π
-2≤1+4k≤4
-3≤4k≤3
-3/4≤k≤3/4
k=0⇒x=π/4
2cos2x+1=0
cos2x=-1/2⇒2x=+-2π/3+2πk⇒x=+-π/3+πk
-π/2≤-π/3+πk≤π
-3≤-2+6k≤6
-1≤6k≤8
-1/6≤k≤8/6
k=0πx=-π/3
k=1⇒x=-π/3+π=2π/3
-π/2≤π/3+πk≤π
-3≤2+6k≤6
-5≤6k≤4
-5/6≤k≤4/6
k=0⇒x=π/3
2
2cos4xcosx-3cos4x=0
cos4x*(2cosx-3)=0
cos4x=0⇒4x=π/2+πk⇒x=π/8+πk/4
0≤π/8+πk/4≤π
0≤1+8k≤8
-1≤8k≤7
-1/8≤k≤7/8
k=0⇒x=π/8
2cosx-3=0
cosx=1,5>1 нет решения