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

In summary: I got the same exact result.In summary, the given equation of cot^2θ+5cosecθ=4 can be simplified to cosec^2θ+5cosec-5=0, which can be solved using the quadratic formula. From the solutions obtained, it is determined that the values of cosecθ are 0.8541 and -5.8541. However, since the range for solutions is 0 °<θ<360 °, the values of θ are converted to degrees and it is found that the solutions in this range are 189.8 ° and 350.16°. It is also noted that the co-functions of sine and tangent, cosecant and cotangent
  • #1
AN630078
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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 👍
 
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  • #2
I got the same result.
 
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  • #3
fresh_42 said:
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°?
 
  • #4
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.
 
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  • #5
What is 'reciprocal' about it?
 
  • #6
epenguin said:
What is 'reciprocal' about it?
Maybe the co's: cosec, cotan; all inverses of the 'normal'.
 
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  • #7
epenguin said:
What is 'reciprocal' about it?
The reciprocal of sinθ is cosecθ.
 
  • #8
fresh_42 said:
Maybe the co's: cosec, cotan; all inverses of the 'normal'.
Yes, that is what I meant 😁
 
  • #9
fresh_42 said:
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?
 
  • #10
Nine point eight degrees something, i.e. ##9.8\ldots °##
 
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  • #11
fresh_42 said:
Nine point eight degrees something, i.e. ##9.8\ldots °##
Oh, ok I was confused by the "something". Thank you for explaining 👍
 
  • #12
AN630078 said:
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 ...:mad:
Not sure whether I can give myself a warning?!
 
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  • #13
fresh_42 said:
My fault, and worst of all: such an internet speak is against the rules ...:mad:
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 😁
 
  • #14
fresh_42 said:
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.
 
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What is a reciprocal trigonometric equation?

A reciprocal trigonometric equation is an equation involving the reciprocal trigonometric functions, such as cotangent, cosecant, and secant. These functions are the inverse of the basic trigonometric functions (sine, cosine, and tangent).

How do I solve a reciprocal trigonometric equation?

To solve a reciprocal trigonometric equation, you can use algebraic manipulation and trigonometric identities to simplify the equation and isolate the variable. You can also use a calculator or a table of values to find the solutions.

What is the cotangent function?

The cotangent function is the reciprocal of the tangent function. It is defined as the ratio of the adjacent side to the opposite side in a right triangle.

What is the cosecant function?

The cosecant function is the reciprocal of the sine function. It is defined as the ratio of the hypotenuse to the opposite side in a right triangle.

How do I check my solutions to a reciprocal trigonometric equation?

You can check your solutions by plugging them back into the original equation and verifying that the equation holds true. You can also graph the equation and see if the solutions intersect with the graph at the given points.

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