Solving equations of the form ##a\sin\theta+b\cos\theta = c##

[brotherbobby]

Homework Statement

Give the general solution to the equation : $\sin\theta+2\cos\theta=1$

Relevant Equations

1. If $\sin\theta = \sin\alpha$ where $\alpha$ is the minimum positive angle whose "$\sin$" is the same as the "$\sin$" of the angle $\theta$, then $\boldsymbol{\theta=n\pi+(-1)^n \alpha}$
2. If $\cos\theta = \cos\alpha$ where $\alpha$ is the minimum positive angle whose "$\cos$" is the same as the "$\cos$" of the angle $\theta$, then $\boldsymbol{\theta=2n\pi\pm \alpha}$
3. $\sin^2\theta = 1-\cos^2\theta$

Problem statement : Given the equation $\sin\theta+2\cos\theta=1$, find the general solution for the angle $\theta$.

Attempt : For the general case where we have $a\sin\theta+b\cos\theta=c$, the line of approach is to take $a=r\cos\alpha$ and $b=r\sin\alpha$ wherein we will have $r=(a^2+b^2)^{1/2}$ and $\alpha= \tan^{-1}\frac{b}{a}$.

For the equation $\sin\theta+2\cos\theta=1$, let $1=r\cos\alpha$ and $2=r\sin\alpha$. These yield $r=\sqrt{5}$ and $\alpha = \tan^{-1}2= 63.43^{\circ}$.
The given equation reduces to $r\sin(\theta+\alpha)=1\Rightarrow \underline{\sin(\theta+\alpha)=\frac{1}{\sqrt{5}}} = \sin\beta\;\text{(say)}$, which gives $\beta = \sin^{-1}(1/\sqrt{5}) = 26.57^{\circ}$.
The underlined equation can be solved using Relevant Equation(1) above to give : $\theta+\alpha = n\pi+(-1)^n\beta\Rightarrow \boxed{\theta = n\pi -\tan^{-1}2+(-1)^n \sin^{-1}(1/\sqrt{5})}$.

Check :
(a) On putting $n=0$, we have $\theta_0 = -\tan^{-1}2+\sin^{-1}(1/\sqrt{5}) = -36.9^{\circ}$ which satisfies the given equation $\large{\checkmark}$.
(b) On putting, $n=1$, we have $\theta_1 = \pi -\tan^{-1}2-\sin^{-1}(1/\sqrt{5}) = 90^{\circ}$ which satisfies the given equation $\large{\checkmark}$.
(c) On putting, $n=2$, we have $\theta_2 = 2\pi -\tan^{-1}2+\sin^{-1}(1/\sqrt{5}) = 323.13^{\circ}$ which satisfies the given equation $\large{\checkmark}$.
[I assume that my general solution above (in box) is correct and will hold for all values of $n \in \mathbb{Z}$.]

The issue : The answer in the book is given as $\boxed{2n\pi+\pi/2\; \text{OR}\; 2n\pi-\gamma}\; \text{where}\; \gamma\; \text{is a positive acute angle whose sine is}\; 3/5$.

Check (Book's answer) :
(a) Putting $n=0$, we have $\theta_0 = 90^{\circ}\;\text{OR}\; \theta_0 = -\gamma = -36.9^{\circ}$. (These match my answers above though for different values of $n$)$\large{\checkmark}$.
(b) Putting $n=1$, we have $\theta_1 = 450^{\circ}\;\text{OR}\; \theta_1 =2\pi -\gamma = 323.1^{\circ}$. (The second one matches my answers above though for a different value of $n$)$\large{\checkmark}$. (The first one may match too for a different value of $n$).

Is my solution correct? I understand that unless I check for several more values of $n$ it is difficult for someone to say.

Alternate solution : My first solution and its correctness is put further into doubt when I realize that the problem can be solved in a different way.

Given $\sin\theta+2\cos\theta=1\Rightarrow \sin^2{\theta}= (1-2\cos\theta)^2\Rightarrow 1-\cos^2\theta=1-4\cos\theta+4\cos^2\theta\Rightarrow 4\cos\theta=5\cos^2\theta$
$\Rightarrow \cos\theta(4-5\cos\theta)=0\Rightarrow \cos\theta = 0=\cos(\pi/2)\;\text{OR} \cos\theta = 4/5=\cos\gamma$
$\Rightarrow \boxed{\theta = 2n\pi+\pi/2}\; \text{OR}\; \boxed{\theta = 2n\pi\pm \cos^{-1}(4/5)}$. This answer is almost like the one in the book with the $\gamma$'s being the same angles. Of course I have a $\pm$ whereas the book only has a $-$sign. I suppose the angles will match since $\cos({-\beta})=\cos{\beta}$.

Question remains - How does the first solution look so different? Is the method alright? A help or suggestion would be welcome.

 

[pasmith]

You know that \cos \theta = \frac45. That means that \sin \theta = 1 - 2 \cos \theta = -\frac35. Now \sin(2n\pi + \alpha) > 0 for 0 < \alpha < \pi, so that's not going to be a solution.

Also: consider the right-angled triangle ABC where AB = 2, BC = 1 and AC = \sqrt{1 + 2^2} = \sqrt{5}. The angle at A has a sine of 1/\sqrt{5} and the angle at C has a tangent of 2. These angles must sum to \pi/2. Thus <br /> \tan^{-1}(2) + \sin^{-1} \left(\frac{1}{\sqrt{5}}\right) = \frac{\pi}{2}. By considering this triangle, you can also show that <br /> \sin\left(\tan^{-1}(2) - \sin^{-1} \left(\frac{1}{\sqrt{5}}\right)\right) = \frac 35 so the two methods do in fact agree.

 

[vela]

Given $\sin\theta+2\cos\theta=1\Rightarrow \sin^2{\theta}= (1-2\cos\theta)^2\Rightarrow 1-\cos^2\theta=1-4\cos\theta+4\cos^2\theta\Rightarrow 4\cos\theta=5\cos^2\theta$
$\Rightarrow \cos\theta(4-5\cos\theta)=0\Rightarrow \cos\theta = 0=\cos(\pi/2)\;\text{OR} \cos\theta = 4/5=\cos\gamma$
$\Rightarrow \boxed{\theta = 2n\pi+\pi/2}\; \text{OR}\; \boxed{\theta = 2n\pi\pm \cos^{-1}(4/5)}$. This answer is almost like the one in the book with the $\gamma$'s being the same angles. Of course I have a $\pm$ whereas the book only has a $-$sign. I suppose the angles will match since $\cos({-\beta})=\cos{\beta}$.

You seemed to have skipped a few steps near the end. Applying your relevant equation #2 to $\cos\theta = 0$ gives $\theta = 2n \pi \pm \pi/2$. Then you need to rule out the solution with the minus sign, which you may have done and just didn't mention it. But I suspected you didn't because for $\cos\theta = 4/5=\cos\gamma$, you included the $\pm$ when solving for $\theta$, but you didn't check if both options work. It turns out the plus sign doesn't as @pasmith noted.