Question

3sin^2 x/2+1/2sinx=2 решите уравнение

Answer 1

$sin^2 x/2=frac{1-cosx}{2}$

получим: 

 

$1,5-1,5cosx+frac{1}{2}sinx=2\3-3cosx+sinx=4\sinx-3cosx=1\(sinx-3cosx)^2=1^2\sin^2x+9cos^2x-6sinxcosx=1\sin^2x+9cos^2x-6sinxcosx=sin^2x+cos^2x\8cos^2x-6sinxcosx=0\4cos^2x-3sinxcosx=0\cosx(4cosx-3sinx)=0\1)cosx=0\x=pi /2+pi k . k=z\2)4cosx-3sinx=0\4cosx/sinx-3sinx/sinx=0/sinx\4ctgx-3=0\ctgx=3/4\x=arcctg(3/4)+pi k . k=z$