anonymous
  • anonymous
i need to solve this over the interval [0,2pi]. tan^2x=2secx-1
Mathematics
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anonymous
  • anonymous
i need to solve this over the interval [0,2pi]. tan^2x=2secx-1
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
tan^2 x - 2 sec x +1
anonymous
  • anonymous
i simplified tan^2x into sec^2x-1. is that a good step?
anonymous
  • anonymous
(sec^2 x -1) - 2 sec x +1 sec^2 x - 2 sec x

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anonymous
  • anonymous
sec (x)=u so substituting u^2-2u=0 We may complete the square by adding 1 to both side u^2 -2u+1=1 (u-1)(u-1)=1 u=2 sec (x)=2 so since 1/sec(x)= cos(x) cos(x)=1/2 x=pi/3
anonymous
  • anonymous
wait what is u?
anonymous
  • anonymous
I just define u for sec(x) for simplicity
anonymous
  • anonymous
ok so is the work with sec(x)=u a continuation of the post you did before that?
anonymous
  • anonymous
U is used in place of sec(x), just to make it easier to see
anonymous
  • anonymous
no, I did not need to use U. I could have instead solve it using sec(x)
anonymous
  • anonymous
Solved without using u sec(x)^2-2sec(x)=0 We may complete the square by adding 1 to both side sec(x)^2 -2sec(x)+1=1 (sec(x)-1)(sec(x)-1)=1 sec(x)=2 sec (x)=2 so since 1/sec(x)= cos(x) cos(x)=1/2 x=pi/3
anonymous
  • anonymous
could i have used the quadratic formula to solve from where it was secx^2-2secx=0?
anonymous
  • anonymous
and also how do you know you have to complete the square?
anonymous
  • anonymous
Yes, but it is much easier
anonymous
  • anonymous
u^2-2u=0 I could see that by adding 1 to left side, I would be able to factor easily. But if you add something one one side, same must be done on the other side
anonymous
  • anonymous
thank you!

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