aviationfan aviationfan
  • 18-03-2015
  • Mathematics
Answered

Solve the equation: sin^2x+2sinx*cosx-3cos^2x=0



Answer :

theresaleigh theresaleigh
  • 19-03-2015
sin²x - 2sinxcosx - 3cos²x = 0 | : cos²x
tg²x - 2tgx - 3 = 0
Substitution : TGX = a
a² - 2a - 3 = 0
by t . vieta : a₁ = 3
 a₂ = -1
inverse substitution :
TGA = 3
 x₁ = arctg3 + piN , n∈Z
TGA = -1
 x₂ = -π / 4 + piN , n∈Z      
On promežutke [ -π , π / 2 ] equation has two roots to -π / 4; arctg3
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