How do you solve $csc^2 x - (cos^4 x + 2 cos^2 x sin^2 x + sin^4 x) = 0$ ?

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Answer 1 · 2016-05-03T23:10:41

$pi/2 + kpi$

First, simplify the expression in parentheses
$cos^4 x + 2 cos^2 x.sin^2 x + sin^4 x =$ (cos^2 x + sin^2 x)^2 = 1^2 = 1#

The equation becomes:
$1/sin^2 x - 1 = 0$
$sin^2 x = 1$
$sin x = +- 1$

a. sin x = 1 --> $x = pi/2 + kpi$
b. sin x = -1 --> $x = (3pi)/2 + kpi$

How do you solve csc^2 x - (cos^4 x + 2 cos^2 x sin^2 x + sin^4 x) = 0 csc^2 x - (cos^4 x + 2 cos^2 x sin^2 x + sin^4 x) = 0?