goniometric equation $\cos(x)-1=3\sin(x)$

Question

I'm having problems solving the equation:

$$\cos(x)-3\sin(x)=1$$

My attempt

$$\begin{align}&\cos(x)= 1+3\sin(x)\\ &\cos(x)-1 = 3\sin(x)& \text{With: }\sin(x)=\sqrt{1-\cos^2(x)}\\ &\cos(x)-1= 3\sqrt{1-\cos^2(x)} \\ &(\cos(x)-1)^2=9-9\cos^2(x)\end{align}$$

Is this right?

Answer 1

Your idea is OK, but I recommend squaring the equation first before using $\sin^2x+\cos^2x=1$ to avoid problems with plus and minus signs (see dxiv's comment). As you have obtained:

$$ \begin{split} (\cos x - 1)^2 &= 9\sin^2x \\ \cos^2x-2\cos x + 1 &= 9-9\cos^2x \\ 10\cos^2x - 2\cos x - 8 &=0 \\ 5\cos^2x - \cos x - 4 &=0\\ (5\cos x + 4)(\cos x -1)&=0 \end{split}$$

Which, gives $\cos x = -4/5$ or $1$. Assuming $x\in[-\pi,\pi)$, then $x=0$ or $\pm \cos^{-1}(-4/5)$.

Edit: checking solutions, we see that $x=\cos^{-1}(-4/5)$ is not a solution. So the two correct solutions are $x=0,-\cos^{-1}(-4/5)$.

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A more direct approach can be using the R-formula:

$$ \sqrt{3^2+1^2}\cos({x-\tan^{-1}(-3)}) = 1 $$

Which gives you $$x=-\tan^{-1}3 \pm \cos^{-1}\left(\frac{1}{\sqrt{10}}\right).$$

This is equivalent to the previous solution, as $\cos^{-1}(1/\sqrt{10})=\tan^{-1}3$ .

Answer 2

Hint: Use the so-called Weierstrass Substitution:

$$\sin(x)=\frac{2t}{1+t^2}$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$ $$\tan(x/2)=t$$

No squaring is needed!

Answer 3

The equation is $-2\sin^2\frac{x}{2}=6\sin\frac{x}{2}\cos\frac{x}{2}$, which is equivalent to at least one of the conditions $\sin\frac{x}{2}=0,\,\tan\frac{x}{2}=-3$ being true. The roots are $2\pi n,\,2\pi n-2\arctan 3$ with integers $n$.

Answer 4

You are on the right track.

$$(\cos(x)-1)^2=\cos^2(x)-2\cos(x)+1=9\sin^2x=9-9\cos^2(x)$$

is a quadratic equation in $\cos(x)$,

$$10\cos^2(x)-2\cos(x)-8=0$$

giving the solutions

$$\cos(x)=\frac{1\pm\sqrt{81}}{10}=1\text{ or }-\frac45.$$ Then using the initial equation,

$$\sin(x)=0\text{ or }-\frac35\text{ respectively}.$$

Now you can draw the possible values of $x$.

$x=2k\pi$ or $x=\arctan\dfrac34+\pi+2k\pi$.

Answer 5

By vector calculus:

$$\cos x-3\sin x=1$$ can be written by means of a dot product

$$(1,-3)\cdot(\cos x,\sin x)=1.$$ Or, after normalization of the first vector,

$$\frac1{\sqrt{10}}(1,-3)\cdot(\cos x,\sin x)=\frac1{\sqrt{10}}=\cos\phi.$$

The fist vector has the direction $-\arctan 3$ and the second direction $x$, and they form an angle $\pm\arccos\dfrac1{\sqrt{10}}$.