Best Answer

6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3

Let f(x) = sinx + 1/3

then the solution to sinx = -1/3 is the zero of f(x)

f'(x) = cosx

Using Newton-Raphson, the solutions are x = 3.4814 and 5.9480

It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.

Q: How do you solve 6sin x 1 plus 9sin x algebraically over the domain 0 x 2pi?

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