Question

Решить уравнение:

sin(x)+cos(x)=1+sin(2x)

Answer 1

(sinx+cosx)^2-(sinx+cosx)=0

sinx=-cosx

tgx=-1

x=-П/4+Пk

sinx-+cosx=1

х=П/2+2Пk

x=2Пk

 

Answer 2

sinx+cosx=1+2sin(2x)

 

sinx+cosx=1+2sinxcosx

sinx+cosx=(sinx)^2+(cosx)^2+2sinxcosx

 

sinx+cosx=(sinx+cosx)^2

1) sinx+cosx=1

$sqrt{2}(frac{sqrt{2}}{2}sinx+frac{sqrt{2}}{2}cosx)=1$

$sqrt{2}(cosfrac{pi}{4}sinx+sinfrac{pi}{4}cosx)=1$

$sqrt{2}sin(x+frac{pi}{4})=1$

$sin(x+frac{pi}{4})=frac{sqrt{2}}{2}$

$x+frac{pi}{4}=frac{pi}{4}+2npi$

$x=(-1)^{n}*(-frac{pi}{4})+npi$

 

2) sinx+cosx=0

sinx=-cosx

sinx/cosx=-1

tgx=-1

x=-Pi/4+Pi*n (n Є N)