0-3+3cos(a/3)=-1/(sin(a/3)) * (9-a)

I have no idea how to do the working out and the answer for a is supposed to be a=5.41

2. Originally Posted by Danie
Solve ... for a

0-3+3cos(a/3)=-1/(sin(a/3)) * (9-a)

... a is supposed to be a=5.41
Rounded to two places, 5.41 is one of the three solutions.

We can simplify the expressions above, by making a substitution.

Let k = a/3 (so a = 3*k)

This substitution leads to an equivalent equation:

1 - cos(k) = (3 - k)/sin(k)

Using an identity for sin(2*k) we also get:

sin(k) - 1/2*sin(2*k) = 3 - k

Graphing both sides of either of these equations shows three solutions for k (see below). Tripling each of these solutions yields the solutions for a.

When we have an equation comprised of a trigonometric function set equal to a polynomial function, it is often impossible to solve by hand.

I approximated the three solutions for k, by zooming in on the intersection points of a graph. One could also use a Computer Algebra System, to approximate the solutions.

There are methods for approximating by hand; have you heard of Newton's Method?

3. I havent heard of Newtons method.

Is there anyway just to end up at 5.41 through algebra only.

Thanks

4. Originally Posted by Danie
Is there anyway just to end up at 5.41 through algebra only.
No. Your equation relates a trigonometric function (transcendental) to a linear function (polynomial). There's no closed-form approach, for finding solutions; we need to approximate the three solutions, using an iterative process of some sort.

difference between polynomial and transcendental equation

5. ok thankyou