 #1
AN630078
 242
 25
 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+5cosecθ4=0
cosec^2θ+5cosec41=0
cosec^2θ+5cosec5=0
Let u=cosecθ
u^2+5u5=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=(53√5)/2=5.8541...
cosecθ=5.8541...
1/sinθ=(53√5)/2
1*2=(53√5)*sinθ
2=(53√5)*sinθ
2/(53√5)=sinθ
Multiply by the conjugate;
2*(5+3√5)/(53√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+5cosecθ4=0
cosec^2θ+5cosec41=0
cosec^2θ+5cosec5=0
Let u=cosecθ
u^2+5u5=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=(53√5)/2=5.8541...
cosecθ=5.8541...
1/sinθ=(53√5)/2
1*2=(53√5)*sinθ
2=(53√5)*sinθ
2/(53√5)=sinθ
Multiply by the conjugate;
2*(5+3√5)/(53√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 👍