How do you solve $2tan^2x - 3secx = 0$ ?

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How do you solve $2tan^2x - 3secx = 0$ ?

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Answer 1 · 2015-07-14T22:53:57

If we restrict the domain of $x$ to $[0, 2pi)$
then $x=(pi)/3 or (5pi)/3$

Explanation:

$2tan^2(x) - 3sec(x) = 0$

$rarr 2(sin^2(x))/(cos^2(x)) - 3(1/cos(x)) = 0$

$color(white)("XXXX")$ assuming $cos(x)!=0$ (the equation would not be valid if it were)
$color(white)("XXXX")$ Multiply by $cos^2(x)$
$2sin^2(x) = 3cos(x)$
$color(white)("XXXX")$ Use Pythagorean relationship
$rarr 2(1-cos^2(x)) -3cos(x) = 0$
$color(white)("XXXX")$ Simplify
$rarr 2cos^2(x)+3cos(x)-2=0$
$color(white)("XXXX")$ Factor
$rarr (2cos(x)-1)(cos(x)+2) = 0$

$rArr 2cos(x)-1 = 0$
or
$color(white)("XXXX")cos(x)+2 = 0$

*Case 1
If $2cos(x)-1 = 0$
$color(white)("XXXX")rarr cos(x) = 1/2$
$color(white)("XXXX")rarr x = (pi)/3 or (5pi)/3$
$color(white)("XXXX")color(white)("XXXX")$ assuming $x in [0,2pi)$

Case 2
If $cos(x)+2 = 0$
$color(white)("XXXX")rarr cos(x) = -2$
BUT
$color(white)("XXXX")cos(x) in [-1,+1] " for all " x$
Therefore this case is extraneous.

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How do you solve $2tan^2x - 3secx = 0$ ?

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