2cosx=sin2x

[Leah5467]

Hello,my question is as above,and my answer is:
(2cosx)=2(sinx)(cosx)
2=2sinx
1=sinx

but there are two answers:

How can i prevent this kind of error happening again?
Thank you!

 

[Harry_the_cat]

You have divided both sides of your equation by cos x. But what if cos x = 0 ?
You can only divide both sides by something that you know is non-zero.

To avoid making that mistake, and losing a solution, it's always better to rearrange the equation to get 0 on one side, factorise, and use the null factor theorem.

 

[Leah5467]

Okay thank you! You have always helped me. Thank you!

 

[tkhunny]

Hello,my question is as above,and my answer is:
(2cosx)=2(sinx)(cosx)
2=2sinx
1=sinx

but there are two answers:
View attachment 12348
How can i prevent this kind of error happening again?
Thank you!

What is [math]2\cos(x) = 2\sin(x)\cos(x)[/math]? -- What "Double Angle Identity" is that?

 

[Leah5467]

is it:0=2sin(x)cos(x)-2(cosx)
0=2cosx(sinx-1)?

 

[tkhunny]

I see. Your reference to "Double Angle Identity" is simply misplaced and has no bearing on this problem statement.

We are to solve: [math]2\cos(x) = 2\sin(x)\cos(x)[/math]
You have manipulated and factored correctly.

[math]0 = 2\cos(x)(\sin(x)-1)[/math]
This is where you have forgotten your algebra - in particular, the Zero Product Property.

If A * B = 0, what is it that we can say about A or B?

 

[Harry_the_cat]

I see. Your reference to "Double Angle Identity" is simply misplaced and has no bearing on this problem statement.

The original question is in the post title, requiring the double angle identity as the first step.