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# Core 2 Help watch

1. Hi, please can someone help me solve the following equation in the interval 0<x<2pi

Cosx(sinx-3cosx)=0

Thanks
2. Posted from TSR Mobile

Multiply out the brackets?
3. (Original post by Hlrk)
Hi, please can someone help me solve the following equation in the interval 0<x<2pi

Cosx(sinx-3cosx)=0

Thanks
What do you get after that...

Posted from TSR Mobile
4. Because it's equal to zero you can solve cosx=0 and (sinx-3cosx)=0 think why this holds true.

Also this thread should be in the maths forum
5. I've multiplied out the brackets and I've managed to get tanx=3cosx. Not sure where to go from there
6. (Original post by Hlrk)
I've multiplied out the brackets and I've managed to get tanx=3cosx. Not sure where to go from there
Firstly ... ... Do not multiply out the brackets

You have a factorised form a(b-c)=0

So, either a=0 or b-c=0
7. (Original post by ttaylor17)
Posted from TSR Mobile

Multiply out the brackets?
There's no need to multiply out the brackets - it would make things less clear here!

If AB = 0 then you know that one (or both) of A and B is 0.
8. So do I just do cos^-1(0)?
9. (Original post by Hlrk)
So do I just do cos^-1(0)?

The bracket gives the others
10. (Original post by Hlrk)
Hi, please can someone help me solve the following equation in the interval 0<x<2pi

Cosx(sinx-3cosx)=0

Thanks
cosx=0 and sinx=3cosx => tanx =1/3
11. (Original post by WatermelonJuice1)
cosx=0 and sinx=3cosx => tanx =1/3
Er, sin x = 3cos x => tan x = 3

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