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

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##.

(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 realise 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
Homework Helper
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 $$\tan^{-1}(2) + \sin^{-1} \left(\frac{1}{\sqrt{5}}\right) = \frac{\pi}{2}.$$ By considering this triangle, you can also show that $$\sin\left(\tan^{-1}(2) - \sin^{-1} \left(\frac{1}{\sqrt{5}}\right)\right) = \frac 35$$ so the two methods do in fact agree.

Last edited:
vela
Staff Emeritus