Question

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

Answer 1

`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`

Answer 2

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`