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# solve 2sin2x = 3sinx watch

2. (Original post by piamiahh)
What have you tried so far?
4. (Original post by piamiahh)
sin2x = 2sinxcosx

You should be able to solve it from there.
5. (Original post by Notnek)
What have you tried so far?
2(2sinxcosx)-3sinx = 0

but don't really know what can happen now
6. (Original post by piamiahh)
2(2sinxcosx)-3sinx = 0

but don't really know what can happen now
So you have

4sin(x)cos(x)-3sin(x) = 0

Notice that sin(x) is a common factor so you can factorise. Try this and post your working if you get stuck.
7. 2sin2x = 3sinx
2(2sinxcosx) - 3sinx = 0

- Started it off for you
8. sin2x=2sinxcosx

This should hopefully be enough.

If you forget this you have the formula sheet which says sin(A+B)=SinACosB+SinBCosA. If you input x as A and B you get sinxcos + sinxcosx which is 2sinxcosx
9. >inb4 divides through by sin(x)
10. (Original post by piamiahh)
2(2sinxcosx)-3sinx = 0

but don't really know what can happen now
what are the common terms between those terms, 2(2sinxcosx) = 3sinx
if you identify it, you can divide both sides from it.
11. 4sinxcosx-3sinx=o.
Sinx(4cosx-3)=0
Sinx=o. Cosx=3/4
12. (Original post by RDKGames)
>inb4 divides through by sin(x)
what are the common terms between those terms, 2(2sinxcosx) = 3sinx
if you identify it, you can divide both sides from it.
13. sinx can be taken out as a factor :-)
14. (Original post by RDKGames)
>inb4 divides through by sin(x)
You lose the sinx solutions if you do that, you’ll only get the ones for cosx unless you meant factor out
15. (Original post by Notnek)
So you have

4sin(x)cos(x)-3sin(x) = 0

Notice that sin(x) is a common factor so you can factorise. Try this and post your working if you get stuck.
got it! thank you!!
16. (Original post by Notnek)
17. (Original post by Zxyn)
You lose the sinx solutions if you do that, you’ll only get the ones for cosx unless you meant factor out
It was a joke.
18. I got sinx = 0 and 4cosx=3
19. cosx = 3 divided by 4

is this correct...

i'm in gcse
20. hope this helps

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