# trig

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How do you solve cos(2arcsin 1/4) using inverse trig. functions??!!

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Compute inverse functions to four significant digits. cos^2x=3-5cosx rewrite it as.. cos^2x + 5cosx -3=0 now you have a quadratic, solve for cos x using the equation cosx=(-5 -+ sqrt (25 +12)/2
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Compute inverse functions to four significant digits. cos^2x=3-5cosx cos^2x + 5cosx -3=0 now you have a quadratic, solve for cos x using the equation cosx=(-5 -+ sqrt (25 +12)/2 the back of the book says .9987+2kpi and -.09987+2kpi …
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