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I missed a couple of leasons and I'm pretty confused, help? Solve the equation[ (sin^2)x - (cos^2)x divided by 1-(sin^2)x] = 5 - tan x

The 'x' in this is an 'x' and not a multiplication symbol

The sin function and cos functions are something you should learn about.

Here is someplace to start (there are tons of math sites, so this may not be best)

http://en.wikipedia.org/wiki/Sine
http://mathworld.wolfram.com/Cosine.html
http://en.wikipedia.org/wiki/Tangent

It should be written like this

[ sin^2(x) - cos^2(x) ] / [ 1 - sin^2(x) ] = 5 - tan(x)

Basically you should use various relationships involving the sine function, the cosine function, and the tangent function to attack the equation algebraically until you can isolate a single expression for sin(x) or tan(x) or cos(x).  At that point, use the appropriate inverse function (arcsin or arccos or arctan, or inverse tangent or inverse sine or inverse cosine) to get specific values for x.  These functions are periodic so an equation like sin(x) = 1/2 will have a primary solution where -pi/2 < x < pi/2 and additional solutions where multiples of 2pi are added and subtracted from it.

http://en.wikipedia.org/wiki/Trigonometry

Also, try not to miss your lessons.


[ sin^2(x) - cos^2(x) ] / [ 1 - sin^2(x) ] = 5 - tan(x)

tan(x) = sin(x)/cos(x)

To reduce typing, use c=cos(x), s=sin(x)

(s^2 - c^2) / (1 - s^2) = 5 - (s/c)

It helps to know s^2 + c^2 = 1 for any x since the point (s,c) is always on the unit circle.

((1-c^2) - c^2) / (c^2) = 5 - s/c

(1 - 2c^2) / (c^2) = 5 - s/c

1/c^2 - 2  = 5 - s/c

sc = 7c^2 - 1

s^2 c^2 = 49 c^4 - 14c^2 + 1

(1-c^2) c^2 = 49 c^4 - 14 c^2 + 1

c^2 - c^4 = 49 c^4 - 14 c^2 + 1

50 c^4 - 15 c^2 + 1 = 0

Solve the quadratic for c^2

Take +/- square root of that to find c

Use inverse cosine to find angle for x