How do you solve #cscx+cotx=1# and find all solutions in the interval #[0,2pi)#?

2 Answers
Noah G
Aug 2, 2016

#1/sinx + cosx/sinx =1#

#(1 + cosx)/sinx = 1/1#

#1+ cosx = sinx#

#(1 + cosx)^2 = (sinx)^2#

#1 + 2cosx + cos^2x = sin^2x#

#1 + 2cosx + cos^2x = 1 - cos^2x#

#2cos^2x + 2cosx + 1 - 1 = 0#

#2cosx(cosx + 1) = 0#

#cosx = 0 and cosx = -1#

#x= pi/2, (3pi)/2, pi#

However, checking in the original equation, you will find that #x = (3pi)/2# is extraneous and #x = pi# makes the equation undefined and is thus also extraneous. Hence, the only solution in the interval #[0, 2pi)# is #{pi/2}#.

Hopefully this helps!

P dilip_k
Aug 2, 2016

#color(red)("solution is "x=pi/2)#

Explanation:

Given equation

#cscx+cotx=1....(1)#

Now we know

#csc^2x-cot^2x=1#

#=>(cscx+cotx)(cscx-cotx)=1#

#=>1*(cscx-cotx)=1#

#=>(cscx-cotx)=1....(2)#

Adding (1) & (2) we get

#2cscx=2=>cscx=1=csc(pi/2)#

#:.x=pi/2#

Again Subtracting (2) from (1)

#2cotx=0=>cotx=0=cot(pi/2)#
#:.x=pi/2#

For #cotx=0=cot(3pi/2)#
then #x=3pi/2# but this does not satisfy the given equation as #sin(3pi/2)=-1#

#color(red)("So only solution is "x=pi/2)#

Alternative

Given equation

#cscx+cotx=1#

#=>1/sinx+cosx/sinx=1#

#=>(1+cosx)/sinx=1#

#=>sinx-cosx=1#

#=>1/sqrt2*sinx-1/sqrt2*cosx=1/sqrt2#

#=>sin(pi/4)*sinx-sin(pi/4)*cosx=1/sqrt2#

#=>sin(x-pi/4)=sin(pi/4)#

#=>x=pi/4+pi/4=pi/2#

Again

#=>sin(x-pi/4)=1/sqrt2=sin(3pi/4)#

#=>x=(3pi)/4+pi/4=pi#
But this does not satisfy the given equation as #cscpi and cot pi" undefined"#

#color(red)("So only solution is "x=pi/2)#

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