Solve $2\cos^2{x}=\sqrt{3}\sin{2x}$.

Question

Problem:

Solve $2\cos^2{x}=\sqrt{3}\sin{2x}$ and give the sum of all the solutions in the interval $0\leq x\leq2\pi.$

Attempt: Using the fact that $\sin{2\theta}= 2\cos{\theta}\sin{\theta}$ on the RHS I get $$2\cos^2{x}=2\sqrt{3}\cos{x}\sin{x}.$$

Dividing by $2\cos{x}$ I get $$\cos{x}=\sqrt{3}\sin{x}.$$

Dividing by $\cos{x}$ again I get $$\tan{x}=\frac{1}{\sqrt{3}} \ \Longleftrightarrow \ x=\pi k+\frac{\pi}{6}, \ \ \forall \in \mathbb{Z.}$$

But it's not correct. Why?

score 6 · Answer 1 · answered Apr 28 '17 at 18:15

6

When you divided by $\cos x$, this assumes $\cos x \ne 0$, but if you have any $x$ with $\cos x = 0$ it is also a solution.

answered Apr 28 '17 at 18:15

gt6989b

54,422

score 3 · Answer 2 · answered Apr 28 '17 at 18:16

3

In the step to obtain $\cos{x}=\sqrt{3}\sin{x}.$ you divides by $\cos(x)$ so you assumed that this quantity was not zero. But $\cos(x)=0$ is an other solution. So you need to solve also this equation.

answered Apr 28 '17 at 18:16

Bérénice

9,367

So, is there any other way of solving this so I avoid dividing by zero? – Parseval Apr 28 '17 at 18:18
1

It is not a problem you just need to say that you consider that $\cos(x)\neq0$ before dividing, and you consider the case $\cos(x)=0$ separately. – Bérénice Apr 28 '17 at 18:19
Wolfram alpha gives me the solutions $x_1 = \frac{\pi}{2}+\pi k$ and $x_2 = \frac{\pi}{6}-\pi k$. I get the oposite signs. – Parseval Apr 28 '17 at 18:22
@A---B: How did you deduce that? – Parseval Apr 28 '17 at 18:23
$\frac{\pi}{6}-\pi k$ is the same as $\frac{\pi}{6}+\pi k$ since $k$ lives in $\mathbb Z$ – Bérénice Apr 28 '17 at 18:23
@Parseval I subtracted $\cos^2 x$ from both sides. – Apr 28 '17 at 18:24
@A---B, Ofcourse! Thanks! – Parseval Apr 28 '17 at 18:24
@Parseval As arnaldo mentioned it should be $\cos x(\sqrt{2}\sin x - \cos x) = 0$, sorry. – Apr 28 '17 at 18:34

score 3 · Accepted Answer · answered Apr 28 '17 at 18:25

The right way to solve is:

$$2\cos^2{x}=2\sqrt{3}\cos{x}\sin{x} \to \cos x(\cos x-\sqrt{3}\sin x)=0$$

what give you:

$$\cos x=0 \text{ or } \cos x-\sqrt{3}\sin x=0$$

For

$$\cos x=0\to x=\frac{\pi}{2}+k\pi$$

and for,

$$\cos x-\sqrt{3}\sin x=0\to \tan x=\frac{\sqrt{3}}{3}\to x=\frac{\pi}{6}+k\pi$$

Solve $2\cos^2{x}=\sqrt{3}\sin{2x}$.

3 Answers3