Expert: Sherman D. - 5/9/2006

Question
I need to solve these three in terms of x

Cos((5pi)/2)+x=-Sin(x)

Cos(x)> -radical(3)/2

Sin(2x)<0

Answer
cos((5pi)/2) + x = -sin(x)
(5pi/2) = 450° or 90°
cos(90°) = 0

x = -sin(x)

x = 0

Check this out, if you go to http://www.calculator.com/calcs/GCalc.html
you know how most trig problems are up and down, if you go to the site above, and type in -(sin(x)) - x, click enter, it will be the wavey form of y = -x

Just make sure the y = part isn't in what you type.

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If by this one you mean

cos(x) > -(sqrt(3))/2
inverse cos both sides
x > 150°

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sin(2x) < 0
2sinxcosx < 0
(sin(x))(cos(x)) < 0

90° < x < 180°
or
270° < x < 360°

Basically, this only works if "x" is in Quadrant II and IV

www.math.com/tables/trig/identities.htm
www.quickmath.com