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August 1, 2014

August 1, 2014

Posted by **sh** on Friday, March 5, 2010 at 3:46am.

- Calculus -
**Reiny**, Friday, March 5, 2010 at 8:00amlet'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

- Calculus -
**drwls**, Friday, March 5, 2010 at 8:03amI 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(2cosx-1) ?

- ???? (something wrong) HELP Calculus -
**Reiny**, Friday, March 5, 2010 at 8:06amJust 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.

- Calculus -
**sh**, Friday, March 5, 2010 at 9:40amI'm pretty sure I typed the question correctly, the answer at the back says 1/2.

- Calculus -
**sh**, Friday, March 5, 2010 at 9:44amRetyped with brackets, it would be

If √2[cosx)-1 = (1+√3)/2 and √2(cosx)+1 = (1-√3)/2, find the value of cos4x.

- Transcription ambiguity -
**MathMate**, Friday, March 5, 2010 at 9:47amIn books where equations are typeset, the square-root sign includes a horizontal bar to indicate the extent of the square-root. 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 square-root includes the "-1".

As they are, there is ambiguity in the given equations.

- Calculus -
**maritza**, Monday, November 22, 2010 at 6:05pmfind the positive roots of the equation

2x^3-5x^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.

- Calculus -
**maritza**, Monday, November 22, 2010 at 6:08pmfind the positive roots of the equation

2x^3-5x^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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