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April 19, 2015

April 19, 2015

Posted by **Kate** on Thursday, May 20, 2010 at 8:35pm.

- Math (trig) -
**Reiny**, Thursday, May 20, 2010 at 10:47pmWhat about it? Are we solving for x ??

if so, then ....

4 cosx + 5 = 6/cosx

4cos^2x + 5cosx = 6

4cos^2x + 5cosx - 6 = 0

(4cosx - 3)(cosx + 2) = 0

cosx = 3/4 or cosx = -2 , Not possible since |cosx| ≤1

so cosx = 3/4

x = 41.4° or x = 221.4°

HELPPPPPPP (TRIG) - 4cos(x) 5=6sec(x)

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