Where does the mark scheme get these numbers from?

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The discussion centers on the derivation of a quadratic equation in a mark scheme, specifically addressing an error in factorization. A user realizes that the quadratic formula was the correct approach rather than the attempted factorization. Another participant points out that the transition from the equation Sin^2(θ) + Sin(θ) - 1 = 0 to Sin(θ)(Sin(θ) - 1) = 0 is incorrect. They suggest distributing the terms to verify the equality. The conversation highlights the importance of accurate mathematical manipulation in solving equations.
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Homework Statement
practicing some Trigonmetric identities and equations questions, and unsure how the mark scheme managed to reach what sin theta is equal to (Q 4B)
Relevant Equations
sin^2 x + cos^2 x =1
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Q4B
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Mark scheme
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My working
 
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NVM guys, i just realised that the quadratic formula was used! Sorry for anyone who may have wasted time
 
Your factorization, 3 lines from the bottom, is incorrect
The step from[Edit]
## Sin^2(\theta)+Sin(\theta)-1=0## , to
##Sin(\theta)[Sin(\theta)-1]=0##
Doesn't follow.
Distribute the terms in the bottom expression to test if they're equal.
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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