# Solve sin(x) = cos(2x) for all x

• Jun 25th 2012, 10:13 AM
Lotte1990
Solve sin(x) = cos(2x) for all x
Hello,

The question is: Solve sin(x) = cos(2x) for all x.

My solution:
sin(x) = cos(2x)
sin(x) = sin(1/2 pi - 2x)
x = 1/2 pi - 2x
3x = 1/2 pi
x = 1/6 pi

However, the question says "solve for all x". I understand that fact that there are multiple x-es since both functions are periodic. But how do I find the other x-es? Please help!
• Jun 25th 2012, 10:18 AM
Prove It
Re: Solve sin(x) = cos(2x) for all x
Quote:

Originally Posted by Lotte1990
Hello,

The question is: Solve sin(x) = cos(2x) for all x.

My solution:
sin(x) = cos(2x)
sin(x) = sin(1/2 pi - 2x)
x = 1/2 pi - 2x
3x = 1/2 pi
x = 1/6 pi

However, the question says "solve for all x". I understand that fact that there are multiple x-es since both functions are periodic. But how do I find the other x-es? Please help!

It would be better to use the identity \displaystyle \begin{align*} \cos{2x} \equiv 1 - 2\sin^2{x} \end{align*}, so that your equation becomes

\displaystyle \begin{align*} \sin{x} &= \cos{2x} \\ \sin{x} &= 1 - 2\sin^2{x} \\ 2\sin^2{x} + \sin{x} - 1 &= 0 \end{align*}

• Jun 25th 2012, 10:49 AM
Lotte1990
Re: Solve sin(x) = cos(2x) for all x
$2\sin^2{x} + \sin{x} - 1 = 0$

$\sin{x} = p$

$2p^2 + p - 1 = 0$

$p^2 + \frac{1}{2}p - \frac{1}{2} = 0$

$p = 1$ or $p = -\frac{1}{2}$

$\sin{x} = 1$ or $\sin{x} = -\frac{1}{2}$

$x = \arcsin{1}$ or $x = \arcsin{-\frac{1}{2}}$

$x = \frac{1}{2}\pi + 2k\pi$ or $x = -\frac{1}{6}\pi + 2k\pi$ or $x = -\frac{5}{6}\pi + 2k\pi$ in which k is an integer.

Is this correct? And is this the COMPLETE answer? Thanks!
• Jun 25th 2012, 12:31 PM
HallsofIvy
Re: Solve sin(x) = cos(2x) for all x
Yes, that is the complete answer.
• Jun 25th 2012, 01:19 PM
Lotte1990
Re: Solve sin(x) = cos(2x) for all x
Ok, then please explain to me why I find x equal to 1/6 pi in the first post, but not in the second one. I don't understand...
• Jun 25th 2012, 04:14 PM
skeeter
Re: Solve sin(x) = cos(2x) for all x
solutions for the interval $[0, 2\pi]$ ...

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

$\sin{x} = 1 - 2\sin^2{x}$

$2\sin^2{x} + \sin{x} - 1 = 0$

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

$\sin{x} = \frac{1}{2}$ ... $x = \frac{\pi}{6} , \frac{5\pi}{6}$

$\sin{x} = -1$ ... $x = \frac{3\pi}{2}$
• Jun 25th 2012, 06:11 PM
Deveno
Re: Solve sin(x) = cos(2x) for all x
Quote:

Originally Posted by Lotte1990
Ok, then please explain to me why I find x equal to 1/6 pi in the first post, but not in the second one. I don't understand...

because you assumed that if sin(x) = sin(pi/2 - 2x), that x = pi/2 - 2x. but this is not true, sin(x) is not a 1-1 function.

for example, sin(pi/2 - 2x) = sin(5pi/2 - 2x) as well (leading to x = 5pi/6), and also cos(2x) = cos(-2x), so

we have sin(x) = sin(pi/2 + 2x) also (leading to x = -pi/2, and thus (by adding 2pi) x = 3pi/2).

it is customary to find the solutions "in one revolution" (either from -pi to pi, or 0 to 2pi), and then add 2kpi to the results.
• Jun 25th 2012, 09:52 PM
Prove It
Re: Solve sin(x) = cos(2x) for all x
Quote:

Originally Posted by Lotte1990
$2\sin^2{x} + \sin{x} - 1 = 0$

$\sin{x} = p$

$2p^2 + p - 1 = 0$

$p^2 + \frac{1}{2}p - \frac{1}{2} = 0$

$p = 1$ or $p = -\frac{1}{2}$

$\sin{x} = 1$ or $\sin{x} = -\frac{1}{2}$

$x = \arcsin{1}$ or $x = \arcsin{-\frac{1}{2}}$

$x = \frac{1}{2}\pi + 2k\pi$ or $x = -\frac{1}{6}\pi + 2k\pi$ or $x = -\frac{5}{6}\pi + 2k\pi$ in which k is an integer.

Is this correct? And is this the COMPLETE answer? Thanks!

You have solved the quadratic equation incorrectly.

\displaystyle \begin{align*} 2p^2 + p - 1 &= 0 \\ 2p^2 + 2p - p - 1 &= 0 \\ 2p(p + 1) -1(p + 1) &= 0 \\ (p + 1)(2p - 1) &= 0 \\ p + 1 = 0 \textrm{ or } 2p - 1 &= 0 \\ p = -1 \textrm{ or } p &= \frac{1}{2} \\ \sin{x} = -1 \textrm{ or }\sin{x} &= \frac{1}{2} \end{align*}