Question

Решить 12. найти сумму корней на промежутке (ctgx+√3)sin2x=0 , на [-100˚; 300˚] 13.(√3ctgx-1)cos(5pi/2-x)=0 , на [-150°; 250°] 14. cos(pi/2+2x)=√2sin(pi+x) ,на[-100˚; 120˚] тоже найти сумму корней на промежутке

Answer 1

12
ctgx=-√3⇒x=5π/6+πn,n∈z
sin2x=0⇒2x=πk⇒x=πk/2,k∈z
x={5π/6;0;π}
5π/6+0+π=11π/6
13
ctgx=1/√3⇒x=π/3+πn,n∈z
cos(5π/2-x)=sinx
sinx=0⇒x=πk,k∈z
x={π/3;4π/3;0;π}
π/3+4π/3+0+π=8π/3
14
cos(π/2+2x)=-sin2x
sin(π+x)=-sinx
sin2x-√2sinx=0
sinx(2cosx-√2)=0
sinx=0⇒x=πn.n∈z
cosx=√2/2⇒x=+-π/4+2πk,k∈z
x={0;-π/4;π/4}
0-π/4+π/4=0

Answer 2

12)
ctgx=-V3⇒x=5π/6+πn,n∈z
sin2x=0⇒2x=πk⇒x=πk/2,k∈z
x={5π/6;0;π}
5π/6+0+π=11π/6
13)
ctgx=1/V3⇒x=π/3+πn,n∈z
cos(5π/2-x)=sinx
sinx=0⇒x=πk,k∈z
x={π/3;4π/3;0;π}
π/3+4π/3+0+π=8π/3
14)
cos(π/2+2x)=-sin2x
sin(π+x)=-sinx
sin2x-V2sinx=0
sinx(2cosx-V2)=0
sinx=0⇒x=πn.n∈z
cosx=V2/2⇒x=+-π/4+2πk,k∈z
x={0;-π/4;π/4}
0-π/4+π/4=0

(V-Корень)