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

1. ## 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:

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

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

2. I found that on the unit circle, i.e. $\displaystyle a^2 + b^2 = 1$, one finds the solution:

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

Is this the only solution to the problem?

3. 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:

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

for $\displaystyle x$. Hence, I want to express $\displaystyle x$ as a function of $\displaystyle a$ and $\displaystyle b$. And for which values of $\displaystyle a$ and $\displaystyle b$ does there exist a solution?
Put $\displaystyle \tan(\theta)=a/b$ then:

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

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

CB