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

2 Answers
Antoine
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

Nghi N.
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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