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I am trying to solve this problem saying, for what values of $a$ and $b$ (in which they are not zero), the equation $a(1-\sin(x))=b(1-\cos(x))$ has any solution(s)? Thank you very much. :)

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up vote 3 down vote accepted

$1- \sin(x) = (\cos(x/2) - \sin(x/2))^2$ and $1 - \cos(x) = 2 \sin^2(x/2)$. This gives us that $$\cot(x/2) - 1 = \pm \sqrt{2b/a}$$ i.e. $$\tan(x/2) = \frac1{1 \pm \sqrt{2b/a}}$$ i.e. $$x = 2 \arctan \left(\frac1{1 \pm \sqrt{2b/a}} \right)$$

Another way to approach is as follows. We have $$a \sin(x) - b \cos(x) = b-a$$ i.e. $$\frac{a \sin(x) - b \cos(x)}{\sqrt{a^2+b^2}} = \frac{b-a}{\sqrt{a^2+b^2}}$$ i.e. $$\sin(x + \phi) = \frac{b-a}{\sqrt{a^2+b^2}}$$ Hence, $$x = \arcsin \left(\frac{b-a}{\sqrt{a^2+b^2}} \right) - \phi$$ where $\phi$ is such that $\displaystyle \cos(\phi) = \frac{a}{\sqrt{a^2+b^2}}$ and $\displaystyle \sin(\phi) = -\frac{b}{\sqrt{a^2+b^2}}$. For $x \in \mathbb{R}$, we need $$-1 \leq \frac{b-a}{\sqrt{a^2+b^2}} \leq 1$$ i.e. $$\frac{(b-a)^2}{a^2+b^2} \leq 1$$ i.e. $$-2ab \leq 0$$ i.e. $$ab \geq 0$$ Hence, $a$ and $b$ must both be of the same sign.

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So, the constants must have the same sign? Thanks Marvis. – Nancy Rutkowskie May 17 '12 at 4:33
@NancyR Yes assuming you want $x \in \mathbb{R}$. – user17762 May 17 '12 at 4:43

An easier way to see that $a$ and $b$ must have the same sign is that $1-\sin x$ and $1-\cos x$ are both positive or zero and they are never zero simultaneously. To show there is a solution for any $a$ and $b$ of the same sign, note that for $a,b \gt 0, a(1-\sin(x))-b(1-\cos(x))$ is negative for $x=\frac \pi 2$ and positive for $x=0$ (reverse this for $a,b \lt 0$). Now use the intermediate value theorem.

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+1. Nice....... – user17762 May 17 '12 at 4:50

Another way to solve it is as follows. From $a(1−\sin(x))=b(1−\cos(x))$ we get


using $1-\sin(x)=(\sin(x/2)-\cos(x/2))^2$ and $1-\cos(x)=2\sin(x/2)^2$, we get


In this case, $0<(b/a)<1/2$ or $(b/a)>1/2$.

So $a$ and $b$ must be the same sign and $a$ can't be $2b$.

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