number of solutions

**adhyeta** · May 3rd 2009, 03:15 AM

let a,b,c be non-zero real numbers such that

$|c|>\sqrt{a^{2}+b^{2}}$

Then find the number of roots of the equation

$asinx + bcosx=c$

???

i was getting no roots as the given expression (on the LHS) according to me satisfies:

$-\sqrt{a^{2}+b^{2}}\leq asinx+bcosx\leq \sqrt{a^{2}+b^{2}}$

which is a totally different set than that of the values of c.

so how can they ever be equal? ???

and the answer is not correct according to the book. :'(

**NonCommAlg** · May 3rd 2009, 03:28 AM

Originally Posted by adhyeta

let a,b,c be non-zero real numbers such that

$|c|>\sqrt{a^{2}+b^{2}}$

Then find the number of roots of the equation

$asinx + bcosx=c$

???

i was getting no roots as the given expression (on the LHS) according to me satisfies:

$-\sqrt{a^{2}+b^{2}}\leq asinx+bcosx\leq \sqrt{a^{2}+b^{2}}$

which is a totally different set than that of the values of c.

so how can they ever be equal? ???

and the answer is not correct according to the book. :'(

if the question is exactly what you gave us here, then your answer is correct and the book is certainly wrong!

**Soroban** · May 3rd 2009, 05:55 AM

Hello, adhyeta!

Let $a,b,c$ be non-zero real numbers such that: . $|c| \;{\color{red}>} \;\sqrt{a^2+b^2}$ . Is this correct?

Then find the number of roots of the equation: . $a\sin x + b\cos x \:=\:c$

Divide by $\sqrt{a^2+b^2}$

. . $\frac{a}{\sqrt{a^2+b^2}}\,\sin x + \frac{b}{\sqrt{a^2+b^2}}\cos x \;=\;\frac{c}{\sqrt{a^2+b^2}}$

Let $\tan\theta \:=\:\frac{b}{a}$
. . Then: . $\cos\theta \:=\:\frac{a}{\sqrt{a^2+b^2}},\;\;\sin\theta \:=\:\frac{b}{\sqrt{a^2+b^2}}$

Then we have: . $\cos\theta\sin x + \sin\theta\cos x \:=\:\frac{c}{\sqrt{a^2+b^2}} \quad\Rightarrow\quad \sin(x + \theta) \:=\:\frac{c}{\sqrt{a^2+b^2}}$ .[1]

But if $|c| \:>\:\sqrt{a^2+b^2}$ , then: . $\frac{|c|}{\sqrt{a^2+b^2}} \:>\:1$

Then [1] becomes: . $\sin(x + \theta) \:> \:1 \quad\hdots$ .which is impossible.

Therefore, there are no solutions.

**adhyeta** · May 3rd 2009, 06:16 AM

is its certainly correct.

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