Question

2sin($frac{7 pi }{2}$-x)*sinx=cosx на [$frac{7 pi }{2}$;5$pi$]

Answer 1

2sin(7π/2-x)*sinx=cosx
-2cosx*sinx-cosx=0
-cosx(2sinx+1)=0
cosx=0⇒x=π/2+πn,n∈z
7π/2≤π/2+πn≤5π
7≤1+2n≤10
6≤2n≤9
3≤n≤4,5
n=3⇒x=π/2+3π=7π/2
n=4πx=π/2+4π=9π/2
sinx=-1/2⇒x=-π/6+2πn U x=-5π/6+2πn
7π/2≤-π/6+2πn≤5π
21≤-1+12n≤30
22≤12n≤31
11/6≤n≤31/12
n=2⇒x=-π/6+4π=23π/24
7π/2≤-5π/6+2πn≤5π
21≤-5+12n≤30
26≤12n≤35
13/6≤n≤35/12
нет решения