设tanx=u,则cosx=土1/√(1+u^2),sinx=tanxcosx=土u/√(1+u^2)方程化为土(u+1)/√(1+u^2)=3-u,平方得(u+1)^2/(1+u^2)=(3-u)^2去分母得u^2+2u+1=(u^2+1)(u^2+2u+1),为4次方程，超出中学数学范围。,
tanX=sinX/cosX（sinX）^2十（cosX）^2=1
