How do you solve $(1/(cos^2 x)) + 1 - 3 tan^2 x = 0$ ?

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Answer 1 · 2015-05-31T00:06:41

$(1/cos^2x)+1−3tan^2x=0$

$sec^2(x)+1-3tan^2x=0$

Since $color(blue)(tan^2x+1=sec^2x)$
then $color(blue)(1= sec^2x-tan^2x)$

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

$2sec^2x-4tan^2x = 0$

Factor and divide out $2$ :

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

$sec^2x-2tan^2x=0$

$1/cos^2x = (2sin^2x)/(cos^2x)$

$1=2sin^2x$

$+-1/sqrt(2) = sinx$

Without restriction:
$x=pi/4 +kpi$

From $0 to 2pi$
$x = pi/4, (3pi)/4, (5pi)/4, (7pi)/4$

You can see it on the graph:

graph{(secx)^2+1-3(tanx)^2 [-10.125, 9.875, -2.28, 7.72]}

How do you solve (1/(cos^2 x)) + 1 - 3 tan^2 x = 0 (1/(cos^2 x)) + 1 - 3 tan^2 x = 0?