# Trigonometric Equations Solution

Hi,

I had to solve without the aid of a calculator:

2cos(x) = sin(2x)

What I did was perform the following substitution:

2cos(x) = 2sin(x)cos(x)

I then cancelled and got:

sin (x) = 1

For 0 to 3pi,

I got two answers, pi/2 and 5pi/2.

I did not get one of the answers. Why?

What did I do wrong?

Thanks!

Related Precalculus Mathematics Homework Help News on Phys.org
SammyS
Staff Emeritus
Homework Helper
Gold Member
Hi,

I had to solve without the aid of a calculator:

2cos(x) = sin(2x)

What I did was perform the following substitution:

2cos(x) = 2sin(x)cos(x)

I then cancelled and got:

sin (x) = 1

For 0 to 3pi,

I got two answers, pi/2 and 5pi/2.

I did not get one of the answers. Why?

What did I do wrong?

Thanks!
2cos(x) = 2sin(x)cos(x) is also true if cos(x) = 0 , (in which case, you divided by zero when you cancelled.)

A safer way:

subtract 2cos(x) from both sides:

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

Factor & use the zero product property of multiplication.

Ok, thanks!

Mark44
Mentor
Just to emphasize what Sammy said, when you're solving equations, it's not a good idea to "cancel" since there is the chance that you will be losing a solution (just like you did here).

Here's a simple example showing why cancelling is not a good idea:

Solve for x in the equation x2 = 4x

First attempt:
Cancel x from each side to get x = 4.
x = 4 is a solution, but the problem is, there is another that was lost by the cancel operation.

Second attempt:
Rewrite the equation as x2 - 4x = 0
Factor to get x(x - 4) = 0
Solution: x = 0 or x = 4