Solving a*cos(x) + b*sin(x) = 1

• Apr 18th 2010, 02:08 PM
sitho
Solving a*cos(x) + b*sin(x) = 1
Hi! The solution to this question may be completely trivial, but I still managed to get stuck on this equation. So I wish to solve:

$a \cos(x) + b \sin(x) = 1$

for $x$. Hence, I want to express $x$ as a function of $a$ and $b$. And for which values of $a$ and $b$ does there exist a solution?
• Apr 18th 2010, 02:37 PM
sitho
I found that on the unit circle, i.e. $a^2 + b^2 = 1$, one finds the solution:

$a = \cos(x)$
$b = \sin(x)$
$\Rightarrow~x = \tan^{-1}\left(\frac{b}{a}\right)$

Is this the only solution to the problem?
• Apr 18th 2010, 10:41 PM
CaptainBlack
Quote:

Originally Posted by sitho
Hi! The solution to this question may be completely trivial, but I still managed to get stuck on this equation. So I wish to solve:

$a \cos(x) + b \sin(x) = 1$

for $x$. Hence, I want to express $x$ as a function of $a$ and $b$. And for which values of $a$ and $b$ does there exist a solution?

Put $\tan(\theta)=a/b$ then:

$a \cos(x)+b \sin(x)=\frac{1}{\sqrt{a^2+b^2}}[\sin(\theta)\cos(x)+\cos(\theta)\sin(x)]=\frac{1}{\sqrt{a^2+b^2}}\sin(x+\theta)=1$

For this to have any real solutions requires that $a^2+b^2 \le 1$.

CB