- #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

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