How do you find all the solutions for $(2 (tan^2)x - 1)((tan^2)x +1) = 0$ ?

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Answer 1 · 2016-06-29T15:09:31

$x=35.26438968^@+-n*180" "$ where $n=0, 1, 2, 3, 4....$

$x=144.7356103172^@+-n*180" "$ where $n=0, 1, 2, 3, 4....$

the given trigonometric equation is

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

using the first factor
$2*tan^2 x-1=0$

$tan^2 x=1/2$

$tan x=+-sqrt(1/2)=+-sqrt2/2$

$x=35.26438968^@+-n*180" "$ where $n=0, 1, 2, 3, 4....$

$x=144.7356103172^@+-n*180" "$ where $n=0, 1, 2, 3, 4....$

using the second factor

$tan^2 x+1=0$

$tan^2 x=-1$

$x= tan^-1 i$

No value for x

God bless....I hope the explanation is useful.

How do you find all the solutions for (2 (tan^2)x - 1)((tan^2)x +1) = 0(2 (tan^2)x - 1)((tan^2)x +1) = 0?