Hi I am a teacher and have a calculus problem that I have a solution to but it seems so involved that I would be interested to see if their were other solutions.

Solve for x, if x is from -90 to 90 degrees

tan2x = 8cos{squared}x - cotx

From Susan
Brampton, Ontario

With some liberal use of the double angle formula for both the sine and the cosine we get four solutions between -pi/2 and pi/2.

Rewrite tan2x = 8 cos^2(x) - cotx as
(sin2x)/(cos2x) = 8 cos^2(x) - (cosx)/(sinx), or
(2 sinx cosx)/(cos2x) = 8 cos^2(x) - (cosx)/(sinx).

If cosx is not zero then dividing by cosx and multiplying by (cos2x)(sinx) the equation becomes
2 sin^2(x) = 8 sinx cosx cos2x - cos2x, or equivalently,

1 - cos2x = 4 sin2x cos2x - cos2x. Thus
1 = 4 sin2x cos2x or
sin4x=1/2.

Therefore possible solutions occur where cosx = 0 or sin4x = 1/4. Checking in the range from -pi/2 to pi/2 gives solutions x= -pi/2, pi/2, pi/24 and (5 pi)/24.

Cheers,
Chris, Denis and Harley