Question

2倍根号3sin2x+4sinx+8倍根号3cosx=17

Answer 1

2√3sin2x+4sinx+8√3cosx=17,设sinx=u,cosx=v,则u^2+v^2=1,①4√3uv+4u+8√3v=17,(4√3v+4)u=17-8√3v,u=(17-8√3v)/(4√3v+4),②代入①,去分母得(17-8√3v)^2+(4√3v^2+4v)^2=(4√3v+4)^2,289-272√3v+192v^2+48v^...