Question

Решить уравнения:

1) 2sin^2x + 3cosx = 0

2) (cos3x-sinx) / sinx = 0

3) 1 - cosx = sin x/2

 

 

 

 

 

 

 

Answer 1

2 sin^x=1-cos^2x =>
2Cos^2x-3cosx-2=0
D=9+4*2*2=25=5^2
X=2
X=-0.5

Answer 2

2sin$^{2}$+cosx+cosx$^{2}$-sinx$^{2}$=0

sin$^{2}$+cosx-sin$^{2}$+1=0

cosx+1=0

cosx=-1

 

(cosx+cosx$^{2}$-sinx$^{2}$-sinx)/sinx=0

(cosx(1+cosx))/sinx-(sinx(sinx+1)/sinx)=0

(cosx(1+cosx))/sinx=-1-sinx

cosx+cosx$^{2}$=-sinx-sinx$^{2}$

cosx$^{2}$+sinx$^{2}$=-cosx-sinx

cosx+sinx=-1

 

2-2cosx=sinx

2-cos$^{2}$-sin$^{2}$=sinx

2=sinx+cosx$^{2}$+sinx$^{2}$

2=sinx+1

sinx=1