質問: How do you find a numerical value of one trigonometric function of x given #(1+cosx)/sinx+sinx/(1+cosx)=4#? ===========================================================================================================
回答: #:.x=pi/6 +2pin or x=(5pi)/6 +2pin# #### Explanation: #(1+cosx)/sinx + sinx/(1+cosx)=4# #((1+cosx)(1+cosx) + sinxsinx)/((1+cosx)sinx)=4# #(1+2cosx+cos^2x+ sin^2x)/((1+cosx)sinx)=4# #(1+2cosx+1)/((1+cosx)sinx)=4# #(2+2cosx)/((1+cosx)sinx)=4# #(2(1+cosx))/((1+cosx)sinx)=4# #(2cancel(1+cosx))/(cancel(1+cosx)sinx)=4# #2/sinx=4# #2=4sinx# #2/4=sinx# #1/2=sinx# #sin^-1(1/2)=x# #:.x=pi/6 +2pin or x=(5pi)/6 +2pin#