Determine all the possible values of x where 0 degree <= x <= 360 degree such that sec^2 x + sec x = 2
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Hey choi105. Can you show us what you have tried?
sec^2 x + sec x = 2 (sec x - 1)( sec x + 2) = 0 sec x - 1 = 0 or sec x + 2 = 0 sec x = 1 or sec x = -2 1/cos x = 1 or 1/cos x = -2 then i don't know how to continue...
There are exact values for +-1/2 and +-1. (Hint Consider pi/2, pi/3 and 2pi/3)
Originally Posted by choi105 sec^2 x + sec x = 2 (sec x - 1)( sec x + 2) = 0 sec x - 1 = 0 or sec x + 2 = 0 sec x = 1 or sec x = -2 1/cos x = 1 or 1/cos x = -2 then i don't know how to continue... Surely you can see where $\displaystyle \displaystyle \cos{(x)} = 1$ and $\displaystyle \displaystyle \cos{(x)} = -\frac{1}{2}$...
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