# Solving Reciprocal Trigonometric Equation cot^2θ+5cosecθ=4

AN630078
Homework Statement:
Hello, I have been practising reciprocal trigonometric equations and have found a question to solve;

cot^2θ+5cosecθ=4 giving all answers in the range 0 °<θ<360 °

I am not actually struggling to solve the equation, but rather how to find the solutions in the given range.
Relevant Equations:
cot^2θ+5cosecθ=4
cot^2θ+5cosecθ=4
cot^2θ+5cosecθ-4=0
cosec^2θ+5cosec-4-1=0
cosec^2θ+5cosec-5=0
Let u=cosecθ
u^2+5u-5=0
u=(-5± 3√5)/2
u=(-5+ 3√5)/2=0.8541...
Substitute cosecθ=u
Therefore, cosecθ=0.8541
1/sinθ=0.8541
sinθ=1/0.8541=1.170... which is not true since sin x cannot be greater than 1 for real solutions.

u=(-5-3√5)/2=-5.8541...
cosecθ=-5.8541...
1/sinθ=(-5-3√5)/2
1*2=(-5-3√5)*sinθ
2=(-5-3√5)*sinθ
2/(-5-3√5)=sinθ
Multiply by the conjugate;
2*(-5+3√5)/(-5-3√5)(-5+3√5)=2*(-5+3√5)/-20=-(-5+3√5)/10
sinθ=-(-5+3√5)/10
My difficulty is solving this last part, do I then take the arcsin of -(-5+3√5)/10?
θ=arcsin(-(-5+3√5)/10)
θ=-0.17166 + 2πn, θ=π+0.17166+2πn
Since the solutions should be in the range 0 °<θ<360 ° should I convert this from radians to degrees?
θ=-0.17166=-9.8353935.. °~-9.84 °which falls outside the specified range, but -9.84 °+360 °=350.16° (which is within the range)
and θ=π+0.17166=3.31325 rad=189.835... ° ~ 189.8 ° to 1.d.p
Other solutions 189.8 °+360 °=549.8° (which lies outside the given range)

So would the solutions in the range 0 °<θ<360 ° be θ=189.8 °and θ=350.16°?
I would be very grateful of any advice 👍

Mentor
2022 Award
I got the same result.

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• AN630078
AN630078
I got the same result.
Thank you for your reply, so the solutions in the range 0 °<θ<360 ° would be θ=189.8 °and θ=350.16°?

Mentor
2022 Award
Yes. I haven't checked the digits behind the point though, but it was 9.8°sth. I checked whether it is a better number in radians, but that didn't look any better. I first thought your polynomial was wrong, until I realized that I calculated in sine instead of your cosecans.

• AN630078
Homework Helper
Gold Member

Mentor
2022 Award
Maybe the co's: cosec, cotan; all inverses of the 'normal'.

• AN630078
AN630078
The reciprocal of sinθ is cosecθ.

AN630078
Maybe the co's: cosec, cotan; all inverses of the 'normal'.
Yes, that is what I meant 😁

AN630078
Yes. I haven't checked the digits behind the point though, but it was 9.8°sth. I checked whether it is a better number in radians, but that didn't look any better. I first thought your polynomial was wrong, until I realized that I calculated in sine instead of your cosecans.

Mentor
2022 Award
Nine point eight degrees something, i.e. ##9.8\ldots °##

• AN630078
AN630078
Nine point eight degrees something, i.e. ##9.8\ldots °##
Oh, ok I was confused by the "something". Thank you for explaining 👍

Mentor
2022 Award
Oh, ok I was confused by the "something". Thank you for explaining 👍
My fault, and worst of all: such an internet speak is against the rules ... Not sure whether I can give myself a warning?!

• AN630078
AN630078
My fault, and worst of all: such an internet speak is against the rules ... Not sure whether I can give myself a warning?!
No do not worry, I am pleased you have used it, now I know its meaning 😁

Homework Helper
Gold Member
Maybe the co's: cosec, cotan; all inverses of the 'normal'.

OK, equations involving the reciprocal trig functions - reciprocal.equations are something else (equations where the reciprocals of roots are also roots).

The co's are not inverses, but the arcsin etc. which you use to get the angles are inverses of the trig functions.

I got the same exact result.
Good insight there that you realized that even if the calculator gives you only one solution, there has to be another.

Last edited:
• AN630078