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# can someone check my trig re-arranging please (C3) watch

1. for all of these trig equation, I have to solve it, but obviously the first thing I have to do is to re-arrange them:

1) 5sin2x + 4sinx = 0

> 10sinxcosx + 4sinx = 0
> 2sinx (5cosx + 2) = 0

2) sin^2 x = 2sin2x

> sin^2 x = 4sinxcosx
> sinx = 4 cosx <--------I took out 'sinx' from both side
> sinx/cosx = 4 <-------- this makes tanx = 4

3) cos^2 x - sin2x = sin^2 x

> cos^2 x - sin^2 x = sin2x <--- I took 'sin^2 x' to the other side
> cos2x = sin2x
> is anyone sure that I am allowed to bring the 'sin2x' to the other side to get: 'cos2x / sin2x = 1'. Seems kinda dodgy since nothing is being multiplied by 'sin2x' in the first place.
2. In (2) you divide both sides by sin(x). That's not OK when sin(x) = 0

So collect the terms on the LHS. Sin(x) will now come out as a factor, and sin(x)=0 will usually lead to some solutions.
3. for question 3--> if you bring over the sin2x you'll have cos2x-sin2x=0
4. In (2) you divide both sides by sin(x). That's not OK when sin(x) = 0
This is correct - never divide an equation through by a variable or you lose a solution. Always factorise out and then solve like ian.slater suggests.

Also for the last part, from cos2x = sin2x going to is perfectly fine - think of the equation as and that might make it easier!
5. be careful in 2, as you have divided both sides by sinx, something which could in fact be zero. can you see how x=0 is actually a solution to 2 initially but after your rearrangement x=0 is no longer a solution? what you really need to do is factor out sinx, so you have (sinx)sinx=(sinx)4cosx and then you can say something like "this is clearly satisfied for sinx=0, i.e., x=2z*pi for some integer z, or otherwise it is satisfied when sinx=4cosx.." and you can then proceed with your argument.
6. fair play. some people are clearly more on the ball than myself. best of luck nonetheless.
7. how does 'sinx = 0' has solutions. I pressed sin inverse times zero, and it comes up with zero, so surely it has no solutions
8. in question 2) as you are dividing by sinx then sinx=0 aswell
9. (Original post by W.H.T)
how does 'sinx = 0' has solutions. I pressed sin inverse times zero, and it comes up with zero, so surely it has no solutions
0 is a solution
10. (Original post by FrancesO)
0 is a solution
so where sould 'sinx = 0' appear on the 'A.S.T.C' graph ??
11. (Original post by W.H.T)
so where sould 'sinx = 0' appear on the 'A.S.T.C' graph ??
The axes count... The graph is cyclical, so although it's 360 when you're going down to C, it's 0 when you're adding up to A
12. (Original post by W.H.T)
how does 'sinx = 0' has solutions. I pressed sin inverse times zero, and it comes up with zero, so surely it has no solutions
sin(0) = 0 so x=0 is a solution of sin(x) = 0.

But trig equations usually have multiple solutions. The question may have asked you to work in degrees or radians. (There are 2pi radians in one complete 'revolution' or 360 deg - introduced in C2 in Edexcel at least)

Usually the question specifies a range for you to find all solutions in. Say 0<= x <= 2pi. You could look at a sine curve to figure that out. Be very careful at the ends of the range is it < or <= ?

After a while you get used to general solutions - so x = n pi satisfies sin(x) = 0 for all integers n.

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