Question

решите уравнение.....​

Answer 1

1) sin(π/2-х)=сosx   , 2cos²x+sin2x=0

2cos²x+2sinx*cosx=0

2cosx(cosx+sinx)=0

cosx=0     или    cosx+sinx=0.

x₁=π/2+πn  или    tgx=-1 , x₂=-π/4+πm

2)a)  3π≤π/2+πn ≤4,5π

3π-π/2≤π/2-π/2+πn ≤4,5π-π/2

2,5π≤πn ≤4π |:π

2,5≤n ≤4 ⇒   n=3,4⇒

x=π/2+π*3=7π/2

x=π/2+π*4=9π/2

б) 3π≤ -π/4+πm ≤4,5π

3π+π/4 ≤-π/4+π/4+πm ≤4,5π+π/4

3,25π ≤ πm ≤4,75π   |:π

3,25≤m ≤4,75   ⇒ m=4⇒  x= -π/4+π*4=15π/4

Ответ 1) π/2+πn  n∈Z;  -π/4+πm  m∈Z

2)7 π/2, 9π/2,15π/4.