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
Solve using the quadratic formula;
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

 

[fresh_42]

I got the same result.

 

[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°?

 

[fresh_42]

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.

 

[epenguin]

What is 'reciprocal' about it?

 

[fresh_42]

What is 'reciprocal' about it?

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

 

[AN630078]

What is 'reciprocal' about it?

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.

Thank you for your reply. Oh splendid, just to ask what do you mean be 9.8°sth?

 

[fresh_42]

Nine point eight degrees something, i.e. $9.8\ldots °$

 

[AN630078]

Nine point eight degrees something, i.e. $9.8\ldots °$

Oh, ok I was confused by the "something". Thank you for explaining

 

[fresh_42]

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]

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

 

[epenguin]

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.