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That equation is impossible to solve because Sin-1(3) is an impossible proportion in a triangle. Earlier in the problem you might have made a mistake. Or, if that is the whole problem you cannot deduce an answer because there is no value of X, Real or imaginary that works in that equation.

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Solve 6sinx equals 1 plus 9sinx algebraically over the domain 0 is greater than or equal to x is less than 2pi?

6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3Let f(x) = sinx + 1/3then the solution to sinx = -1/3 is the zero of f(x)f'(x) = cosxUsing Newton-Raphson, the solutions are x = 3.4814 and 5.9480It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.


Sinx plus cosx equals 0?

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

6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3Let f(x) = sinx + 1/3then the solution to sinx = -1/3 is the zero of f(x)f'(x) = cosxUsing Newton-Raphson, the solutions are x = 3.4814 and 5.9480It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.


How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?

(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx


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Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?

2


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Is there any way to solve a system of equations with C and D as constants and x and y as variables sinx plus cozy - C equals 0 cosx plus siny - D equals 0?

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How do you solve csc x-sin x equals cos x cot x?

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