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Okay, will keep my question as straightforward as possible:

Solve the following equation algebraically over the interval 0 is equal to or less than x < 2pi and round the answer to the nearest hundredth of a radian.

The equation is: csc^(2)x=5

Thank you for your help.

Hi John,

The easy way to do this would be by using a Maclaurin series expansion.

We have

csc²(x) = 5

csc(x) = √5

and

sin(x) = 1/√5 ≈ ±0.4472

Now,

sin(x) = x - (x^3)/6 + (x^5)/120 + .......

The function, x - (x^3)/6 + (x^5)/120, approximates sin(x) very well in the interval 0 < x < π/2 but x - (x^3)/6 also does well and is easier to solve so we'll use it.

Solving for the positive value,

x - (x^3)/6 = 0.4472

which gives x ≈ 0.464 or just 0.46 and lies in the required interval.

We do not need to solve for the negative value since the function is odd and we know that x = -0.46 would also be a solution but is outside our interval. We can always use the solution that lies in the interval 0 < x < π/2 to find the other solutions.

In general, for

sin(A) = B

where 0 < A < π/2 and B > 0 (as it has to be), then

sin(π - A) = B

sin(π + A) = -B

sin(2π - A) = -B

And so the complete solution to sin(x) = ±0.4472 would be

x = 0.464, π - 0.464, π + 0.464, 2π - 0.464

and to the nearest hundredth

x = 0.46, 2.68, 3.61, 5.82

and are consequently the approximate solutions to csc²(x) = 5

We could also have used the Maclaurin series for csc(x) directly, since

csc(x) = (1/x) + x/6 - 7x³/360 + ...

or just (1/x) + x/6 in the interval 0 < x < π/2 as before.

I do not know what you're studying exactly but I see here that it's in calculus so i supposed that how to solve a cubic equation isn't the issue here and have therefore skipped the details of that. You can always get back to me if anything is unclear.

Regards

Calculus

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