ShirokovP

Verified answer

Sin^2x + 2sinxcosx - (cos^2x + sin^2x) = 0 
sin^2x + 2sinxcosx - cos^2x - sin^2x = 0 
cosx (2sinx - cosx) = 0 

1) cosx =  0
x = pi/2 + pik, k ∈Z

2) 2sinx - cosx = 0   /:cosx≠ 0
2tgx - 1 = 0 
tgx = 1/2
x = arctg(1/2) + pik, k ∈Z

Dимасuk

Verified answer

Sin²x + sin2x = 1
Разложим синус удвоенного аргумента:
sin²x + 2sinxcosx - 1 = 0
-(1 - sin²x) + 2sinxcosx = 0
-cos²x + 2sinxcosx = 0
cosx(-cosx + 2sinx) = 0

cosx = 0
x = π/2 + πn, n ∈ Z

cosx = 2sinx
1 = 2tgx
tgx = 1/2
x = arctg(1/2) + πn, n ∈ Z.