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(2cosx-1) ?
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 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.
find 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.