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November 30, 2015
Posted by **tatiana** on Thursday, October 27, 2011 at 3:35pm.

- pre-calculus -
**Steve**, Thursday, October 27, 2011 at 5:17pmSince cos(2x) = 1 at x = 0,pi,2pi,3pi,...

sec(x) = 1 at 0,2pi,4pi,6pi,...

They meet at even multiples of pi.

cos never gets above 1, sec never gets below 1.

On your interval, x=0 is the only solution

Analytically,

cos 2x = sec x

2cos^2 x - 1 = 1/cosx

2 cos^3 x - cos x - 1 = 0

cosx = 1 is the only real solution.

so, x=0

- pre-calculus -
**pana**, Thursday, October 27, 2011 at 5:49pmcos(2theta)= 120/169; 2theta in QIV