How do you solve #1 = cot^2 x + csc x#?

2 Answers
Jul 22, 2015

#x=(-1)^k(-pi/6)+kpi#

for #k in ZZ#

Explanation:

#cot^2x+cscx=1#

Use the identity : #cos^2x+sin^2x=1#

#=>cot^2x+1=csc^2x#

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

Substitute this in the original equation,

#csc^2x-1+cscx=1#

#=>csc^2x+cscx-2=0#

This a quadratic equation in the variable #cscx# So You can apply the quadratic formula,

#csx=(-1+-sqrt(1+8))/2#

#=>cscx=(-1+-3)/2#

Case #(1) :#
#cscx=(-1+3)/2=1#

Rememeber that : #cscx=1/sinx#

#=>1/sin(x)=1=>sin(x)=1=>x=pi/2#

General solution (1) : #x=(-1)^n(pi/2)+npi#

We have to reject(neglect) these values because the #cot# function is not defined for multiples of #pi/2# !

Case #(2) :#
#cscx=(-1-3)/2=-2#

#=>1/sin(x)=-2=>sin(x)=-1/2=>x=-pi/6#

General solution (2) : #x=(-1)^k(-pi/6)+kpi#

Jul 22, 2015

Solve cot^2 x + csc x = 1

Ans: #(pi)/2; (7pi)/6 and (11pi)/6#

Explanation:

#cos^2 x/sin^2 x + 1/sin x = 1#
#cos^2 x + sin x = sin^2 x#
#(1 - sin^2 x) + sin x = sin^2 x#
#2sin^2 x - sin x - 1 = 0 --> 2t^2 - t - 1 = 0# - Call sin x = t

Since a + b + c = 0, use shortcut: 2 real roots are :
t = 1 and #t = -1/2#
a. t = sin x = 1 --> #x = pi/2#
b. #sin x = - 1/2# --> #x = (7pi)/6# and #x = (11pi)/6#

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