Expert: Sherman D. - 12/22/2006

Question
Solve the equation for X is between 0 and 360 degress.  square root of 2 cosX + 1 =0
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The text above is a follow-up to ...

-----Question-----
I need help simplifying the equation: (sinX + cos X)^2 +(sinX-cosX)^2.  THanks!!!
-----Answer-----
(sin(x) + cos(x))^2 + (sin(x) - cos(x))^2

((sin(x) + cos(x))(sin(x) + cos(x)) + ((sin(x) - cos(x))(sin(x) - cos(x))

(sin(x)^2 + sin(x)cos(x) + sin(x)cos(x) + cos(x)^2) + (sin(x)^2 - sin(x)cos(x) - sin(x)cos(x) + cos(x)^2)

((sin(x)^2 + cos(x)^2) + 2sin(x)cos(x)) + ((sin(x)^2 + cos(x)^2) - 2sin(x)cos(x))

(1 + sin(2x)) + (1 - sin(2x))

1 + sin(2x) + 1 - sin(2x)

ANS : 2

Info found at
http://www.math.com/tables/trig/identities.htm

also if you go to www.quickmath.com, click on Simplify, and type your problem in the way i did, it will give you

2(cos(x)^2 + sin(x)^2)

which we all know that cos(x)^2 + sin(x)^2 = 1, so the true answer would be just 2.

Answer
Even though this isn't really a follow-up question, i will answer it.

square root of 2 cosX + 1 =0

If by this you mean

sqrt(2cos(x) + 1) = 0
2cos(x + 1) = 0
cos(x + 1) = 0
x + 1 = 90° or 270°
ANS : x = 89° or 269°

Are you sure you don't mean

cos(x)^2 + 1 = 0
whereas cos(x) is the only thing being squared.