Having trouble solving Sin2x-1=0? Let's tackle it together!

[Determined77]

Ok this is my first post...so bear with me My problem is:

Sin2x-1=0

I thought I recognized that Sin2x is from the double angle identities chapter so I substituted [2CosSin] for Sin2x.

So I ended up with 2CosSin=1 ( I moved the one over)

...and then I got stuck... am I attacking this wrong?

 

[cristo]

You don't need to use the double angle formulae here. You have the equation sin(2x)=1.. try taking sin^-1 of both sides.

 

[Determined77]

Sorry if I'm dense but I'm not sure what happens after I do that, or what it looks like...

If I took sin-1 from both sides would I get 2x=1 and
Sin-1(1) or 90 degrees?

 

[cristo]

Sorry, by "take sin^-1 of both sides" I didn't mean subtract it (besides, sin^-1 without an argument is an operation, not a function).

Anyway, I meant do this: sin^-1{sin(2x)}=sin^-1(1).

As you rightly said, sin^-1(1)=90 degrees (strictly, it is equal to 360*n+90) degrees, where n=0,1,... but I don't think this is required here).

What does the left hand side, sin^-1{sin(2x)}, become?

 

[Gib Z]

What you do to one side, you must do to the other!

\sin(2x)=1

\arcsin(\sin(2x)) = \arcsin 1

2x= \arcsin (1)

x = \frac{\arcsin 1}{2}

 

[HallsofIvy]

By the way, you titled this "Solving Trig Identities" so one thing that may be confusing you is that this is NOT an identity (an equation that is true for all x). It is an equation that is true for some x and not for others. The problem is to solve the equation. You do that, as both cristo and Gib Z said, by "doing the opposite". Since the equation is sin(2x)= 1, you get rid of the sin by "doing the opposite"- the inverse sine (or "arcsin") :
arcsin(sin(2x))= arcsin(1) ==> 2x= arcsin(1)
(Other notation for the same thing:sin^-1(sin(2x))= sin^-1(1))

Now that you have 2x= arcsin(1) you get rid of the "2" by again "doing the opposite"- the opposite of multiplying by 2 is dividing by 2:
2x/x= x= arcsin(1)/2 or x= sin^-1(1)/2.

If you want a specific decimal approximation to that answer, your calculator should have a "sin^-1(x)" function right next to the "sin" function.

 

[Gib Z]

Note arcsin 1 = \pi/2 + 2k\pi , k \in \mathbb{Z}

 

[HallsofIvy]

Note arcsin 1 = \pi/2 + 2k\pi , k \in \mathbb{Z}

Well, if you want to do the easy way!

 

[f(x)]

I thought I recognized that Sin2x is from the double angle identities chapter so I substituted [2CosSin] for Sin2x.

So I ended up with 2CosSin=1 ( I moved the one over)

...and then I got stuck... am I attacking this wrong?

Its much easier to get the values if the problem stays in one trig func.

 

[cristo]

Its much easier to get the values if the problem stays in one trig func.

Did you not read every other reply in this thread?...