#### BYUguy

Hi,

PROBLEM: "Find all values of x such that sin2x = sinx and 0 ≤ x ≤ 2π" (that last symbol is "Pi" in case it's hard to read)

ANSWERS: 0, π/3, π, 5π/3, 2π

MY THOUGHT PROCESS BEFORE SEEING THE ANSWER WAS:

(1) double-angle formula: sin2x = (2)(sinx)(cosx)
(2) sin2x = sinx only when (2)(cosx) = 1
(3) (2)(cosx) = 1 only when cosx = 1/2
(4) cosx = 1/2 only when theta = π/3

Yes, π/3 is an answer, yet there are more answers (0, π, 5π/3, 2π). So, what was wrong with my thought process, and why did it lead me to only one out of the 5 answers?

And what would be a better way to approach this problem, or a way to fix my initial approach?

Thanks.

Last edited:

#### skeeter

MHF Helper
$\sin(2x)=\sin{x}$

$2\sin{x}\cos{x}=\sin{x}$

$2\sin{x}\cos{x}-\sin{x}=0$

$\sin{x}(2\cos{x}-1)=0$

set each factor equal to zero and solve ...

$\sin{x}=0$ at $x=0$, $x=\pi$, and $x=2\pi$

$2\cos{x}-1=0 \implies \cos{x}=\dfrac{1}{2}$.

Last two solutions are $x=\dfrac{\pi}{3}$ and $x=\dfrac{5\pi}{3}$

• 1 person

#### BYUguy

Oh, yeah, now I see that even the way I was looking at it should have given me 5π/3 right away. And actually the other three as well.

Thanks!

#### Prove It

MHF Helper
(2) sin2x = sinx only when (2)(cosx) = 1
THIS is your mistake. The cardinal rule of numbers is that you can not divide by 0. As Skeeter has shown you, it is possible that sin(x) = 0 (in fact, it's where the other solutions come from).

ALWAYS solve equations by using the Null Factor Law if possible.

• 1 person

#### BYUguy

Thanks to both of you!