1
$\begingroup$

I need to solve the equation $$\sqrt{3}\sin(x)+\cos(x)-2=0$$

My try: I separated the radical then I squared and I noted $\cos(x)=t$ and I got a quadratic equation with $t=\frac{1}{2}$

To solve $\cos(x)=\frac{1}{2}$ I used formula $x\in\left \{ +-\arccos(\frac{1}{2})+2k\pi \right \}k\in\mathbb{Z}$ and I got $x\in\left \{ -+ \frac{\pi}{3}+2k\pi \right \}$ but the right answer is $x=(-1)^{k}\cdot\frac{\pi}{2}+k\pi-\frac{\pi}{6}$

Where's my mistake?How to get the right answer?

$\endgroup$
  • $\begingroup$ ]Hint this can be written as $$\frac{\sqrt{3}}{2}\sin(x)+\frac{1}{2}\cos(x)=1.$$ and $$\left(\frac{\sqrt 3}2,\frac 12\right)=\left(\sin \frac{\pi}{3},\cos\frac{\pi}3\right)$$ $\endgroup$ – Thomas Andrews May 10 at 21:54
  • $\begingroup$ The way you write the "right answer" is pretty cumbersome imo.: why not better use $\;2k\pi\;$ and etc.? Look the answer below $\endgroup$ – DonAntonio May 10 at 21:58
  • $\begingroup$ You "separated the radical then squared". I presume you got $\sqrt{3}\sin(x)= 2- \cos(x)$ and then squared to get $3sin^2(x)= 3- 3cos^2(x)= 4- 4cos(x)+ \sin^2(x)$. And setting "cos(x)= t", $3- 3t^2= 4- 4t+ t^2$ or $4t^2- 4t- 3= 0$. No, t= cos(x)= 1/2 does not satisfy that. $\endgroup$ – user247327 May 10 at 22:09
  • 1
    $\begingroup$ it's $4t^2-4t+1=0$ $\endgroup$ – DaniVaja May 10 at 22:12
  • $\begingroup$ @DaniVaja Your mistake is that you got a solution for original equation before squaring of $\sqrt{3}\sin(x) = 2 - \cos(x)$, but also for $-\sqrt{3}\sin(x) = 2 - \cos(x)$. I explain in more detail in my answer. $\endgroup$ – John Omielan May 11 at 4:29
5
$\begingroup$

I like the following way.

By C-S $$2=\sqrt3\sin{x}+\cos{x}\leq\sqrt{((\sqrt3)^2+1^2)(\sin^2x+\cos^2x)}=2.$$ The equality occurs for $$(\sqrt3,1)||(\sin{x},\cos{x})$$ or since $$\sqrt3\sin{x}+\cos{x}\geq0,$$ for $$\sin{x}=\frac{\sqrt3}{2}$$ and $$\cos{x}=\frac{1}{2},$$ which gives $$x=60^{\circ}+360^{\circ}k,$$ where $k\in\mathbb Z$.

$\endgroup$
  • 2
    $\begingroup$ I really like answers that use inequalities in order to solve these equations! +1 $\endgroup$ – Dr. Mathva May 10 at 22:40
4
$\begingroup$

An idea:

$$\sqrt3\,\sin x+\cos x-2=0\implies\frac{\sqrt3}2\sin x+\frac12\cos x=1\iff\sin\left(x+\frac\pi6\right)=1\implies$$

$$\implies x+\frac\pi6=\frac\pi2+2k\pi\implies x=\frac\pi3+2k\pi$$

$\endgroup$
2
$\begingroup$

Your mistake was forgetting that when you square a non-$0$ equality, the result satisfies two possible equations. In particular, both $x = y$ and $x = -y$ gives that $x^2 = y^2$. From what you describe, you did the following:

$$\sqrt{3}\sin(x) = 2 - \cos(x) \tag{1}\label{eq1}$$

then square both sides to get

$$3\sin^2(x) = 4 - 4\cos(x) + \cos^2(x) \tag{2}\label{eq2}$$

Next, you made certain manipulations to get

$3 - 3\cos^2(x) = 4 - 4\cos(x) + \cos^2(x) \; \Rightarrow \; 4\cos^2(x) - 4\cos(x) + 1 = 0$

Using $\cos(x) = t$ then gives

$$4t^2 - 4t + 1 = 0 \; \Rightarrow \; (2t - 1)^2 = 0 \; \Rightarrow \; t = \frac{1}{2} \tag{3}\label{eq3}$$

The full set of solutions for $\cos(x) = \frac{1}{2}$ is $x = \pm\frac{\pi}{3} + 2k\pi, k \in \mathbb{Z}$. However, note the RHS of \eqref{eq1} is always $\frac{3}{2}$, but for $x = \frac{\pi}{3} + 2k\pi$ the LHS is also $\frac{3}{2}$, but for $x = -\frac{\pi}{3} + 2k\pi$ the LHS is $-\frac{3}{2}$ instead. Squaring with either case still gives \eqref{eq2}.

Whenever you use a non-reversible operation, like squaring, it's important you check to remove any extraneous results you may have got from your manipulations. However, checking by substituting your results into the original equation is generally always a good idea in case you made some mistake.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.