I'm having problems solving the equation:


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?

  • $\begingroup$ Uhm... Since when is $\sin x= \sqrt{1-\cos^2 x}$? That's never been a thing, AFAIK. $\endgroup$ – Saucy O'Path Aug 25 '18 at 6:47
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    $\begingroup$ Yes, your idea is right, but you have to remember that squaring an equation can introduce additional roots that do not necessarily satisfy the original equation. $\endgroup$ – dxiv Aug 25 '18 at 6:50
  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ – José Carlos Santos Aug 25 '18 at 6:52
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    $\begingroup$ @SaucyO'Path: you are kidding, hopefully. $\endgroup$ – Yves Daoust Aug 25 '18 at 7:49
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    $\begingroup$ @YvesDaoust As in, I should have reminded him that $$\sin(x)=\begin{cases}\sqrt{1-\cos^2 x}&\text{if }\exists k\in\Bbb Z,\ 0\le x-2k\pi\le \pi\\ -\sqrt{1-\cos^2 x}&\text{if }\exists k\in\Bbb Z,\ -\pi< x-2k\pi<0\end{cases}\quad ?$$ $\endgroup$ – Saucy O'Path Aug 25 '18 at 7:58

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!

  • $\begingroup$ Except the squaring of $t$ ;-) $\endgroup$ – Yves Daoust Aug 25 '18 at 8:07

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 $n$.

  • $\begingroup$ Nice observation. $\endgroup$ – Yves Daoust Aug 25 '18 at 12:15

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 $x=0,-\cos^{-1}(-4/5)$.

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 $\cos^{-1}(1/\sqrt{10})=\tan^{-1}3$ .

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    $\begingroup$ You must check your solutions, since you have squared! $\endgroup$ – Dr. Sonnhard Graubner Aug 25 '18 at 7:39
  • $\begingroup$ Thank you, I have amended the solution. $\endgroup$ – YiFan Aug 25 '18 at 7:50

You are on the right track.


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


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


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 $\pm\arccos\dfrac1{\sqrt{10}}$.


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