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To three decimal places, find the value of the first positive x-intercept for the function f(x) = 2cos(x + 3).
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I got -1.429 but i think that's wrong
it has to be wrong, since it is not positive
\[2\cos(x+2)=0\\ \cos(x+2)=0\\ x+2=\frac{\pi}{2}\] i guess that gives you the negative answer right?
try \[x+2=\frac{5\pi}{2}\]
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