Kuenzie M.
asked • 07/24/19Find all solutions of the equation
Find all solutions of the equation in the interval [0,2π)
=sec2x/3 +sqrt2=0
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1 Expert Answer
If sec((2x/3)) = -sqrt(2) then 1/cos((2x/3)) = - 1/sqrt(2) and cos((2x/3)) = - 707
Then 2x/3 = cos-1(-.707) or 2x/3 = 135 & 225 (Solved for 2x/3 is in degrees)
So, x = (3/2)135 and (3/2)225
x = 202.5 and 337.5 but since cos(-x) = cos(x) , the x values can be positive negative
So, x = +- 202.5, +- 337.5
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Mark H.
Is that supposed to be (sec^2 (x) ) / 3? (in prose: "find the sec of x, square it, and then divide by 3") Or---could it be the secant of the expression 2x/3 ..? Or---the square of the secant of x/3 Add parentheses as needed to clarify.....07/24/19