Question

2sin2x-sinxcosx=cos2x

Answer 1

Ответ:

⊆$x=\left \{ {{arctan(\frac{-3+\sqrt{13} }{2} )+k\pi } \atop {-arctan\frac{3+\sqrt13}{2}}+k\pi } \right.$

Объяснение:

2*2sinxcosx-sinxcosx=cosx^2-sinx^2

4sinxcosx-sinxcosx=cosx^2-sinx^2

3sinxcosx-sinxcosx-cosx^2+sinx^2=0

3sinxcosx-(1-sinx^2)+sinx^2=0

3sinxcosx-1+sinx^2+sinx^2=0

3sinxcosx-1+2sinx^2=0

3sinxcosx+2sinx^2=1

3sinxcosx+2sinx^2=sinx^2+cosx^2

3sinxcosx+2sinx^2-sinx^2-cosx^2=0

3sinxcosx+sinx^2-cosx^2=0

3tanx+tanx^2-1=0

3t+t^2-1=0

t=$\frac{-3+\sqrt{13} }{2}$

t=$\frac{-3-\sqrt{13} }{2}$