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I'm trying to solve: $\sin 2x=2x^2-4$.
I am quite familiar with the trigonometric identities, and the methods for solving trigonometric equations where all the expressions involved are trigonometric functions (for example: $\sin x\cos x=\tan 2x$) or equations that involve both trigonometric functions and constants, but I just can't remember the way to solve an equation of the above form.

I am not interested in the solution, but rather need someone to refresh my memory on how to handle an equation such like the above mentioned.

Thanks.

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    $\begingroup$ For starters, you know there can only be a real solution if $-1\leq 2x^2-4\leq 1$. $\endgroup$ – MPW Apr 26 '14 at 12:27
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As re0 wrote, this kind of equations, which mix trigonometric functions and polynomials, cannot be solved analytically and only numerical methods can be used.

If you plot the function, you should notice that the equation has two roots, one close to $-1.5$ and the other close to $1.5$ (take care, the curve is not symmetric - the numbers I gave are just rough approximations).

SInce we have reasonable guesses, we can use Newton iterations which, for solving $f(x)=0$, write $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, for your problem, $$f(x)=\sin 2x-2x^2+4$$ $$f'(x)=2\cos 2x-4x$$ Let us start for the negative root at $x_0=-1.5$. The successive iterates will then be $-1.34052$, $-1.32976$, $-1.32972$ which is the solution for six significant figures.

I let you doing the same for the positive root.

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The exact solution of such equation cannot be obtained. But they can be approximated graphically. We can draw the graphs of the trigonometric and polynomials and find the approximate points where they intersect.

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  • $\begingroup$ How can I get an approximation? can I use Taylor series for it? Is every equation of that form can only be approximated? $\endgroup$ – so.very.tired Apr 26 '14 at 12:27
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    $\begingroup$ @so.very.tired Newton's method is a common numerical root finding algorithm. And yes, roots to equations of this type almost always can only be approximated. $\endgroup$ – David H Apr 26 '14 at 12:34
  • $\begingroup$ OK thanks for the help. $\endgroup$ – so.very.tired Apr 26 '14 at 12:38

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