I have a simple equation which i cannot solve for $x$:
$$A\cos x + B\sin x = C$$
Could anyone show me how to solve this. Is this a quadratic equation?
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See this answer for the general approach and an example. – Ayman Hourieh Jul 16 '13 at 10:12
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My picture-answer to a related question may be helpful. – Blue Jul 16 '13 at 11:19
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$A\cos x+B\sin x=C$ so if $A\neq 0, B\neq 0$ then $$\frac{A}{\sqrt{A^2+B^2}}\cos x+\frac{B}{\sqrt{A^2+B^2}}\sin x=\frac{C}{\sqrt{A^2+B^2}}$$ in which $$\frac{A}{\sqrt{A^2+B^2}}\le1,~~\frac{B}{\sqrt{A^2+B^2}}\le1,~~\frac{C}{\sqrt{A^2+B^2}}\le1$$ This means you can suppose there is a $\xi$ such that $\cos(\xi)=\frac{A}{\sqrt{A^2+B^2}},\sin(\xi)=\frac{B}{\sqrt{A^2+B^2}}$ and so...
Mikasa
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+1 . I think the condition needed here is merely $,A\neq 0;;or;;B\neq 0;$ , though if one of them is zero then the equation is much easier. – DonAntonio Jul 16 '13 at 10:18
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@DonAntonio: Yes dear Don. It is my wrong using of notation. I'll fix. Sorry for being Lazy in typing. – Mikasa Jul 16 '13 at 10:20
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1@71GA: I don't know :) But there is a great comment above under question. Follow that. – Mikasa Jul 16 '13 at 10:53
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HINT:
We can also utilize Weierstrass substitution (1, 2), which will convert the given equation to a Quadratic equation in $\tan \frac x2$
lab bhattacharjee
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