sec2x-1=0 [0, 2pi) solve for x

Trent Carpenter 2021-09-14 Answered
\(\displaystyle{\sec{{2}}}{x}-{1}={0}{\left[{0},{2}\pi\right)}\)
solve for x

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

Demi-Leigh Barrera
Answered 2021-09-15 Author has 15692 answers
\(\displaystyle{\left(\frac{{1}}{{\cos{{2}}}}{x}\right)}-{1}={0}\)
\(\displaystyle{\left(\frac{{1}}{{\cos{{2}}}}{x}\right)}-{\left({\cos{{2}}}\frac{{x}}{{\cos{{2}}}}{x}\right)}={0}\)
\(\displaystyle\frac{{{1}-{\cos{{2}}}{x}}}{{\cos{{2}}}}{x}={0}\)
\(\displaystyle{1}-{\cos{{2}}}{x}={0}\)
\(\displaystyle{1}-{\left({\left({\cos{{x}}}\right)}^{{2}}-{\left({\sin{{x}}}\right)}^{{2}}\right)}={0}\)
\(\displaystyle{1}-{\left({\cos{{x}}}\right)}^{{2}}+{\left({\sin{{x}}}\right)}^{{2}}={0}\)
\(\displaystyle{\left({\sin{{x}}}\right)}^{{2}}+{\left({\sin{{x}}}\right)}^{{2}}={0}\)
\(\displaystyle{2}{\left({\sin{{x}}}\right)}^{{2}}={0}\)
\(\displaystyle{\left({\sin{{x}}}\right)}^{{2}}={0}\)
\(\displaystyle{\sin{{x}}}={0}\)
\(\displaystyle{x}={0}\) or \(\displaystyle{x}=\pi\)
Have a similar question?
Ask An Expert
15
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question
...