Find the positive value of $\displaystyle x$ which satisfies $\displaystyle x=3.3cos(x)$, accurate to 5 decimal places. I know how to use Newton's Method, I just can't figure this one out since there's nothing equal to zero.
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Originally Posted by john11235 Find the positive value of $\displaystyle x$ which satisfies $\displaystyle x=3.3cos(x)$, accurate to 5 decimal places. I know how to use Newton's Method, I just can't figure this one out since there's nothing equal to zero. $\displaystyle f(x)=3.3 \cos(x) - x$ CB
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