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

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

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Answer 1 · 2016-05-03T23:10:41

$\frac{π}{2} + k π$ $pi/2 + kpi$

Explanation:

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

The equation becomes:
$\frac{1}{{sin}^{2} x} - 1 = 0$ $1/sin^2 x - 1 = 0$
${sin}^{2} x = 1$ $sin^2 x = 1$
$sin x = \pm 1$ $sin x = +- 1$

a. sin x = 1 --> $x = \frac{π}{2} + k π$ $x = pi/2 + kpi$
b. sin x = -1 --> $x = \frac{3 π}{2} + k π$ $x = (3pi)/2 + kpi$

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

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How do you solve csc2x−(cos4x+2cos2xsin2x+sin4x)=0 csc^2 x - (cos^4 x + 2 cos^2 x sin^2 x + sin^4 x) = 0?

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