How do you solve $Sin(pi/2 - x) cot(pi/2 + x) = -sin x$ from 0 to 2pi?

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How do you solve $Sin(pi/2 - x) cot(pi/2 + x) = -sin x$ from 0 to 2pi?

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Answer 1 · 2016-03-15T00:46:27

undefined equation.

Explanation:

sin (pi/2 - x) = cos x (supplementary arc)
$cot (pi/2 + x) = (cos (pi/2 + x))/(sin (pi/2 + x)) = sin x/(-cos x).$
The left side of the equation simplifies to:
$(cos x)(sin x/(-cos x)) = - sin x$ ,
Then, the equation reduces to:
-sin x = - sin x
The equation is undefined. It is insolvable.

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How do you solve $Sin(pi/2 - x) cot(pi/2 + x) = -sin x$ from 0 to 2pi?

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How do you solve $Sin(pi/2 - x) cot(pi/2 + x) = -sin x$ from 0 to 2pi?