2sec(3x-(pi/3)) Find the period.

Am I right on this? 2sec^2(2x)-8=0 2sec^2(2x)=8 divide by 2 sec^2(2x)=4 find square r...

f(x)= 2sec(x-3.14/4)-1

tan^2(x)=-3/2sec(x)

2sec(11)sec(19)-2cot(71)

2sec^2(2x+5) dx what's the answer when dx is .3 it keeps saying i'm wrong with the one i've found...

If tan A=a/b prove that 2sec theta + 1/ cos theta+2=Õ a2+b2/b

2sec^2x -3tanx =5 for [0,2pi) Please solve for the values of x for me.

2sec^2x - 3tanx - 5 = 0 over the interval 0 to 2pi x= 1.14, 4.28, 5.68, and 2.54 right?

Sec^2(u/2)=(2sec(u))/(sec(u)+1)

t=2sec, s=2m,v=?

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