Calculus
posted by sh .
If √2cos(x)1 = (1+√3)/2 and √2cos(x)+1 = (1√3)/2, find the value of cos4x.

let's add the two given equations
√2cos(x)1 = (1+√3)/2
√2cos(x)+1 = (1√3)/2
2√2cosx = 1
cosx = 1/(2√2) or 1/√8
Then by Pythagoras , sinx = √7/√8
using some identities
cos 4x = cos^2 2x  sin^2 2x
= (cos 2x)(cos 2x)  (sin 2x)(sin 2x)
= (cos^2 x  sin^2 x)^2  (2sinxcosx)^2
= (1/8  7/8)^2  (2√7/8)^2
= 36/64  28/64
= 8/64
= 1/8 
I don't "get" it.
If √2cos(x)1 = (1+√3)/2,
you can solve for x directly, and then calculate 4x, and its cosine. There may be more than one answer. The second equation may have a different set of answers with some agreeing with the first set.
You need to clarify what is "under" the square root sign, using parentheses. Is it [sqrt2*(cosx)] 1 or sqrt(2cosx1) ? 
Just looking over this question again
especially the first equation
√2cos(x)1 = (1+√3)/2
√2cos(x) = (1+√3)/2 + 1
√2cosx = (3 + √3)/2
cosx = (3+√3)/(2√2) = 1.673
but the cosine of any angle is between 1 and +1,
so it would be undefined.
check the typing of the question please. 
I'm pretty sure I typed the question correctly, the answer at the back says 1/2.

Retyped with brackets, it would be
If √2[cosx)1 = (1+√3)/2 and √2(cosx)+1 = (1√3)/2, find the value of cos4x. 
In books where equations are typeset, the squareroot sign includes a horizontal bar to indicate the extent of the squareroot. When the equation is transcribed to a single line, the horizontal line should be replaced with parentheses.
For example:
√2cos(x)1 should be written as √(2)cos(x)1 if the horizontal line is short, and √(2cos(x))1 if the horizontal line extends over the cos(x), and √(2cos(x)1) if the squareroot includes the "1".
As they are, there is ambiguity in the given equations. 
find the positive roots of the equation
2x^35x^2+6=0
and evaluate the function y=500x^6 at each root. round you answers to the nearest integer, but only in the final step.
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