Asked by Alex
Am I right on this?
2sec^2(2x)-8=0
2sec^2(2x)=8
divide by 2
sec^2(2x)=4
find square root
sec(2x)=2
Now, would this simply be asking me to find where secant is equal to one? Or is there no solution because secan't (pun intended) be equal to anything above 1?
2sec^2(2x)-8=0
2sec^2(2x)=8
divide by 2
sec^2(2x)=4
find square root
sec(2x)=2
Now, would this simply be asking me to find where secant is equal to one? Or is there no solution because secan't (pun intended) be equal to anything above 1?
Answers
Answered by
Damon
sec^2(2x)=4
find square root
sec 2x = 2
or
sec 2x = -2
find square root
sec 2x = 2
or
sec 2x = -2
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