Wednesday

August 20, 2014

August 20, 2014

Posted by **jacob** on Monday, April 30, 2007 at 10:33pm.

cos^3(2x) + 3cos^2(2x) + 3cos(2x) = 4

I havent done trig for a while so what exactly does that mean in solving for x on that interval and how would i go about doing that?

Where are you getting these tough trig questions??

how about letting cos 2x = t

then your equation becomes

t^3 + 3t^2 + 3t - 4 = 0

using Newton's Method I was able to come up with ONE real root , t=.709976

so cos 2x = .709976

for 2x= .7813322 radians

and finally by dividing by 2

x = appr. .39

You can factorize the cubic equation Reiny gave above:

t^3 + 3t^2 + 3t - 4 = 0 --->

(t+1)^3 = 5 --->

t = -1 + 5^(1/3)

Count, now that was clever.

good for you to recognize the sequence

1,3,3,(1) in Pascal's triangle.

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