I really don't know, can someone help me...

simple algebra:

sin(2x) = \frac{1}{2}

2x = sin^{-1}(\frac{1}{2})

x = \frac{sin^{-1}(\frac{1}{2})}{2}

This will give you the lowest value of x. You need to remember that there are in fact an infinite number of values for x, as the graph at the bopttom of the post shows. Each of my lines shows the positions of solutions to sin(2x) = 1/2 between 0 and 10. Working them out should be fairly trivial.

Let me know if you need more help!

simple algebra:

sin(2x) = \frac{1}{2}

2x = sin^{-1}(\frac{1}{2})

x = \frac{sin^{-1}(\frac{1}{2})}{2}

This will give you the lowest value of x. You need to remember that there are in fact an infinite number of values for x, as the graph at the bopttom of the post shows. Each of my lines shows the positions of solutions to sin(2x) = 1/2 between 0 and 10. Working them out should be fairly trivial.

Let me know if you need more help!

you insane?? You can't multiply or divide with sin!!

this is the solution

sin(2x)=1/2

{sin(30°)=1/2}

sin(2x)=sin(30°)
2x=30°
x=15°

Graphic side is OK, but nul points aren't asigned so I can only assume what are they

you insane??? you can't multiply or divide with sin!!!

this is the solution

sin(2x)=1/2

{sin(30°)=1/2}

sin(2x)=sin(30°)
2x=30°
x=15°

Hey, be polite at least!:mad: Some people are trying to help, and they may have their own methods. You don't have to criticize other's methods, since you yourself couldn't solve the problem earlier.

I wasn't dividing by sin, i was using the inverse sin operator on both sides, where sin^{-1}(sin(x)) = x. This is perfectly acceptable, and all scientific calculators have this function.

Also, the graph wasn't for you to read off the answer, it was to show you that there are an infinite number of values of 2x. Your answer gives only one of them (and i point out that your solution is the same as mine!).

Hey, be polite at least!:mad: Some people are trying to help, and they may have their own methods. You don't have to criticize other's methods, since you yourself couldn't solve the problem earlier.

True... my apologies, I bumped into the solution after some time...

I wasn't dividing by sin, i was using the inverse sin operator on both sides, where sin^{-1}(sin(x)) = x. This is perfectly acceptable, and all scientific calculators have this function.

Also, the graph wasn't for you to read off the answer, it was to show you that there are an infinite number of values of 2x. Your answer gives only one of them (and i point out that your solution is the same as mine!).

so you're saying that [ sin(x) / sinx ] = x? Don't know man, this is the fist time i see smth like this...

[ sin(x) / sin ] =x

sorry, not used to write sin all alone...

you used the arc function... sin^(-1) = arc



I used the same function in my 3rd step...
both our solutions are correct...

and I would like to apologise to you for calling you insane, sorry ,)...

generally, ^(-1) of a function implies the inverse of the function. For example if f(x) = 2(x^2) then f^-1(x) = (1/2x)^0.5. not simply 1/f(x)

You need to recognise the other solutions, x=pi/12 isn't the only answer. You don't have a complete solution yet.