How do you solve #sin^2 x - sin x = 1# over the interval #[0, 2pi]# ?

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Answer 1 · 2018-06-04T02:41:50

#x = pi - text{Arc}text{sin}( 1/2(1-sqrt{5})) # or

#x = 2pi + text{Arc}text{sin}( 1/2(1-sqrt{5}))#

I think I just did this one a couple of days ago.

#sin^2 x - sin x - 1 = 0#

#sin x = 1/2 (1 pm \sqrt{5})#

Only the negative sign gives a real #x#.

#sin x = 1/2(1 - sqrt{5})#

There will be a pair of supplementary angles with this sine in the range.

#x = arcsin( 1/2(1-sqrt{5})) #

In the range that's

#x = pi - text{Arc}text{sin}( 1/2(1-sqrt{5})) # or

#x = 2pi + text{Arc}text{sin}( 1/2(1-sqrt{5}))#

That's around #322^circ# and #218^circ# but I'll leave the calculator stuff to you.

Interestingly, half the Golden Ratio is #cos 36^circ# but we have to whole Goldren Ratio here.