Thread: Identities and Letting Variables equal

1. Identities and Letting Variables equal

I'm having trouble figuring out how to solve cos x = 2sinxcosx within 0 to 2pi

I know that 2sinxcosx is an identity for sin2x but that doesn't really help the situation since I want to let cosx be a variable (ie. let cosx = m and then solve for roots/angles)

I was thinking about factoring the second side of the equation but it seems like I can't get anything out of that either, I would just get

cos x = sinx (2cosx)
or
cos x = cosx (2sinx)

which doesn't really do anything Anyways if anyone can help solve it by letting either cosx or sinx being a variable it would greatly be appreciated The answer is supposedly 3pi/2, thanks for the help in advance!

2. Originally Posted by peekay
I'm having trouble figuring out how to solve cos x = 2sinxcosx within 0 to 2pi

I know that 2sinxcosx is an identity for sin2x but that doesn't really help the situation since I want to let cosx be a variable (ie. let cosx = m and then solve for roots/angles)

I was thinking about factoring the second side of the equation but it seems like I can't get anything out of that either, I would just get

cos x = sinx (2cosx)
or
cos x = cosx (2sinx)

which doesn't really do anything Anyways if anyone can help solve it by letting either cosx or sinx being a variable it would greatly be appreciated The answer is supposedly 3pi/2, thanks for the help in advance!

sin(2x) doesn't come into use here, this is a factorising question

Method 1

Take cos(x) from both sides:

$2sin(x)cos(x) - cos(x) = 0$

Factor out cos(x):

$cos(x)(2sin(x)-1) = 0$

Either $cos(x) = 0 \rightarrow x = \frac{\pi}{2}$

Or $sin(x) = \frac{1}{2} \rightarrow x = \frac{\pi}{6}$

Since cos(x) is symmetrical about $\pi$ then $\frac{\pi}{2} + \pi = \frac{3\pi}{2}$ is also a solution

Also note that because 2sin(x) is symmetrical about pi/2 then it follows that there is another root at $\frac{\pi}{2}+({\pi}{2} - \frac{\pi}{6}) = \frac{5\pi}{6}$.

Therefore the equation 2sin(x)cos(x) = cos(x) has roots of $\frac{\pi}{6} \: , \: \frac{\pi}{2} \: , \: \frac{5\pi}{6} \: and \: \frac{3\pi}{2}$

---------

Method 2

Divide through by cos(x)

$2sin(x) = 1$

$sin(x) = \frac{1}{2}$

$x = \frac{\pi}{6} \: , \: \frac{5\pi}{6}$

Note that method 1 is better because it yielded two extra solutions

3. Thanks for both the solutions, really do appreciate it.

I like method 2 cause you can just spot it out right away, but considering I'll be writing an exam on monday I better try and get all the solutions I can get .

4. Originally Posted by e^(i*pi)
sin(2x) doesn't come into use here, this is a factorising question

Method 1

Take cos(x) from both sides:

$2sin(x)cos(x) - cos(x) = 0$

Factor out cos(x):

$cos(x)(2sin(x)-1) = 0$

Either $cos(x) = 0 \rightarrow x = \frac{\pi}{2}$

Or $sin(x) = \frac{1}{2} \rightarrow x = \frac{\pi}{6}$

Since cos(x) is symmetrical about $\pi$ then $\frac{\pi}{2} + \pi = \frac{3\pi}{2}$ is also a solution

Also note that because 2sin(x) is symmetrical about pi/2 then it follows that there is another root at $\frac{\pi}{2}+({\pi}{2} - \frac{\pi}{6}) = \frac{5\pi}{6}$.

Therefore the equation 2sin(x)cos(x) = cos(x) has roots of $\frac{\pi}{6} \: , \: \frac{\pi}{2} \: , \: \frac{5\pi}{6} \: and \: \frac{3\pi}{2}$

---------

Method 2

Divide through by cos(x)

$2sin(x) = 1$

$sin(x) = \frac{1}{2}$

$x = \frac{\pi}{6} \: , \: \frac{5\pi}{6}$

Note that method 1 is better because it yielded two extra solutions
Method 2 is not valid.

You're dividing by a quantity that potentially can equal zero. So not only are you comitting an invalid operation, you're losing (as you point out yourself) potential solutions.

$\cos (x) = \sin (2x) \Rightarrow \sin\left( \frac{\pi}{2} - x \right) = \sin (2x)$.
Case 1: $\frac{\pi}{2} - x = 2x + 2n \pi \Rightarrow x = \frac{\pi}{6} - \frac{2n \pi}{3}$ where n is an integer.
Case 2: $\frac{\pi}{2} - x = (\pi - 2x) + 2n \pi \Rightarrow x = (2n + 1) \pi - \frac{\pi}{2}$ where n is an integer.