How do you find all solutions of the equations $sec^2x-1=0$ in the interval $[0,2pi)$ ?

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How do you find all solutions of the equations $sec^2x-1=0$ in the interval $[0,2pi)$ ?

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Answer 1 · 2017-06-17T19:38:18

$0,pi$

Explanation:

$"using "$

$• secx=1/cosx$

$rArr1/cos^2x-1=0$

$rArrcos^2x=1$

$rArrcosx=+-1$

$cosx=1rArrx=cos^-1(1)=0$

$cosx=-1rArrx=cos^-1(-1)=pi$

$rArrx=0,pito[0,2pi)$

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How do you find all solutions of the equations $sec^2x-1=0$ in the interval $[0,2pi)$ ?

1 Answer

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How do you find all solutions of the equations sec^2x-1=0sec^2x-1=0 in the interval [0,2pi)[0,2pi)?

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How do you find all solutions of the equations $sec^2x-1=0$ in the interval $[0,2pi)$ ?