解方程sinx+cosx+tanx+cotx+secx+cscx+2=0求详解哈

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解方程sinx+cosx+tanx+cotx+secx+cscx+2=0求详解哈
解方程sinx+cosx+tanx+cotx+secx+cscx+2=0
求详解哈

解方程sinx+cosx+tanx+cotx+secx+cscx+2=0求详解哈
sinx+cosx+(sinx/cosx)+(cosx/sinx)+1/sinx+1/cosx+1+1=0
(sinx+1)(1+1/cosx)+(cosx+1)(1+1/sinx)=0
(sinx+1)(1+cosx)/cosx=-(cosx+1)(sinx+1)/sinx
又六个三角函数都要有意义,故sinx不为0,1,-1
gu cosx=-sinx
x=k(pie)-(pie)/4

sinx+cosx+tanx+cotx+secx+cscx+2=0
(sinx+cosx)+1/(sinxcosx)+(cosx+sinx)/(sinxcosx)+2=0
(sinx+cosx)(1+1/(sinxcosx))+(1+1/(sinxcosx))+1=0
(sinx+cosx+1)(1+1/(sinxcosx))=-1
(sinx+cosx+1)(1+...

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sinx+cosx+tanx+cotx+secx+cscx+2=0
(sinx+cosx)+1/(sinxcosx)+(cosx+sinx)/(sinxcosx)+2=0
(sinx+cosx)(1+1/(sinxcosx))+(1+1/(sinxcosx))+1=0
(sinx+cosx+1)(1+1/(sinxcosx))=-1
(sinx+cosx+1)(1+sinxcosx)=-sinxcosx
设u=sinx+cosx sinxcosx=(u^2-1)/2
(u+1)(u^2+1)/2=-(u^2-1)/2
(u+1)(u^2+1)=(1-u^2)
u^2+1=1-u
u^2+u=0
u(u+1)=0
u=0或u=-1
sinx+cosx=√2sin(x+45)
√2sin(x+π/4)=0 或 √2sin(x+π/4)=-1
x=-π/4+kπ x=-π/2 +2kπ

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sinx+cosx+tanx+cotx+secx+cscx+2=0
(sinx+cosx+1)(1+sinxcosx)=-sinxcosx
设u=sinx+cosx sinxcosx=(u^2-1)/2
(u+1)(u^2+1)/2=-(u^2-1)/2
(u+1)(u^2+1)=(1-u^2)
u^2+1=1-u
u=0或u=-1
sinx+cosx=√2sin(x+45)
√2sin(x+π/4)=0 或 √2sin(x+π/4)=-1
x=-π/4+kπ x=-π/2 +2kπ