Asked by tatiana
solve cos 2x=secx for 0(<=)x<2pi using trigonometric identities.
Answers
Answered by
Steve
Since 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
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
Answered by
pana
cos(2theta)= 120/169; 2theta in QIV
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