C3 cot, cosec etc iterative formula

can anyone solve:

solve for 0<y<360

2cot^2y + 5cosecy + cosec^2y = 0

plus this one:

use the iterative formula

xn+1=2+ 1/((xn)^2)+1)

with a suitable starting value to find the root of the eqn f(x)=0 correct to 3sf and justify the accuracy of your answer.

what value should i start with?

Reply 1

AlphaNumeric

Try the substituion cot²x + 1 = cosec²x and solve the resulting quadratic in cosec.

Reply 2

Dekota

jenh1

solve for 0<y<360
2cot^2y + 5cosecy + cosec^2y = 0

2/tan²y + 5/siny + 1/sin²y = 0
2cos²y/sin²y + 5/siny + 1/sin²y = 0
2cos²y + 5siny + 1 = 0
2(1-sin²y) + 5siny + 1 = 0
2 - 2sin²y + 5siny + 1 = 0
2sin²y - 5siny - 3 = 0

Let s = siny, then ..
2s² - 5s - 3 = 0
s² - 5/2s - 3/2 = 0
(s - 5/4)² - 25/16 - 3/2 = 0
(s - 5/4)² - 49/16 = 0
s = 5/4 ± √(49/16)
s = 5/4 ± 7/4

sub s = siny = 12/4 ...

siny = 12/4
y = arcsin(3), no solutions.

sub s = siny = -2/4 ...

siny = -½
y = arcsin(-½), y = -30° etc.
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jenh1

xn+1=2+ 1/((xn)^2)+1)

with a suitable starting value to find the root of the eqn f(x)=0 correct to 3sf and justify the accuracy of your answer.

what value should i start with?

1) sketch the graph, and find an interval for x.
2) Do a decimal search to find an interval for x.

Reply 3

steve2005

jenh1

Use the iterative formula

xn+1=2+ 1/((xn)^2)+1)

with a suitable starting value to find the root of the eqn f(x)=0 correct to 3sf and justify the accuracy of your answer.

what value should i start with?

In my head I put x = 1 which gave a new x of 2.5 so I started with 2.5