设tan(x/2)=u,由万能公式,sinx=2u/(1+u^2),cosx=(1-u^2)/(1+u^2),原式变为(1+u^2)/(2u)+4(1+u^2)/(1-u^2)=7*2u(1-u^2)/(1+u^2)^2,去分母得(1-u^2+8u)(1+u^2)^3=28[u(1-u^2)]^2,8次方程,繁!设cosx+4sinx=m,①则cosx=m...