document.write( "Question 390443: How do you algebraically solve x(sinx) = cosx ? \n" ); document.write( "

Algebra.Com's Answer #276851 by richard1234(7193) $\"\"$ $\"About$

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Not very easy to solve this algebraically, since trigonometric functions are transcendental (meaning \"not algebraic\"), and most equations that combine transcendental and algebraic functions cannot be solved algebraically. You could approximate the solutions graphically, or use a technique called Newton's method, which uses some introductory calculus.\r
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\n" ); document.write( "\n" ); document.write( "Newton's method approximates the zeros of a function via a recursive sequence. Move all terms to one side to get $\"x+sin+x+-+cos+x+=+0\"$ , and let $\"f%28x%29+=+x+sin+x+-+cos+x\"$ . Pick an initial guess and denote it $\"x%5B0%5D\"$ . Then, for $\"n+%3E=+0\"$ ,\r
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\n" ); document.write( "\n" ); document.write( " $\"x%5Bn%2B1%5D+=+x%5Bn%5D+-+f%28x%5Bn%5D%29%2F%28f%3A%28x%5Bn%5D%29%29\"$ (The colon represents an apostrophe, the parsing system doesn't accept f'(x[n])).\r
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\n" ); document.write( "\n" ); document.write( "Note that f'(x[n]) is the derivative of the function at x[n].\r
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\n" ); document.write( "\n" ); document.write( "Repeating this process, $\"x%5Bn%5D\"$ converges to one of the zeros of f(x).\r
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