Sin^2x+1/sin^2x если sinx-1/sinx=-3

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вопрос

Given:

sinx - 1/sinx = -3

solving for sin^2x:

(sin^2x - 1) / sinx = -3
sin^2x - 1 = -3sinx
sin^2x + 3sinx - 1 = 0

Using the quadratic formula, we have:

sinx = [-3 +- sqrt((3)^2 - 4(1)(-1))] / 2
sinx = [-3 +- sqrt(9 + 4)] / 2
sinx = [-3 +- sqrt(13)] / 2

Therefore, sinx = (-3 + sqrt(13)) / 2 or sinx = (-3 - sqrt(13)) / 2

Now, to find sin^2x + 1/sin^2x:

sin^2x + 1/sin^2x = (sin^2x)^2 / sin^2x + 1/sin^2x
= sin^4x / sin^2x + 1/sin^2x
= sin^2x + 1/sin^2x
= ((-3 + sqrt(13)) / 2)^2 + 2 / (-3 + sqrt(13))
= ((-3 + sqrt(13))^2 + 4) / 4

This is the final result for sin^2x + 1/sin^2x when sinx - 1/sinx = -3.