1/sin(x)+4/cos(x)=7*sin(x)cos(x)x为0到180度闭区间求x

Question

1/sin(x)+4/cos(x)=7*sin(x)cos(x) x为0到180度闭区间 求x

Answer 1

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-4sinx,②
(cosx)^2+(sinx)^2=m^2-8msinx+17(sinx)^2=1,
17(sinx)^2-8msinx+m^2-1=0,
sinx=[4m土√(17-m^2)]/17,
①^2,1+8sinxcosx+15(sinx)^2=m^2,
∴sinxcosx=[m^2-1-15(sinx)^2]/8,
原式变为m=7(sinxcosx)^2,也繁!
设sinx+cosx=t,则sinxcosx=(t^2-1)/2,
(sinx-cosx)^2=1-2sinxcosx=2-t^2,
sinx-cosx=土√(2-t^2),
cosx+4sinx=(5/2)(sinx+cosx)+(3/2)(sinx-cosx)=5t/2土(3/2)√(2-t^2),
原方程变为5t/2土(3/2)√(2-t^2)=7[(t^2-1)/2]^2,繁!
尚无捷径!