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Asked by: david0287-ga
List Price: $5.00
29 Apr 2006 09:52 PDT
Expires: 29 Apr 2006 12:03 PDT
Question ID: 723943
I have read a few websites and a book on trigonometric identities and I know that: cos x = sin 90 - x cos^2 + sin^2 x = 1 sin x / cos x = tan x I am not sure how to solve an equation whith z(sin x) and (cos y) For example sovle the equation: 3(sin 20) = cos(2x) [degrees] (find x for 0 <= x <=180) Thanks
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From: kottekoe-ga on 29 Apr 2006 11:50 PDT
In general, equations like this cannot be solved without invoking the so-called inverse trigonometric functions. For example, to solve your example for x: z*sin(y) = cos(2x) We could write: 2x = arccos(z*sin(y)), where arccos(a) is the inverse of the cosine function Thus, we can formally solve it by saying: x = arccos(z*sin(y))/2 Unfortunately, your example has no solution, since 3*sin(20) is bigger than one and thus cannot equal cos(2x) for any value of x.
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