Prep Calc Exam

[CrystalTennis]

2sin²x=1 where 0 is less than or equal to x and x is less than 2 pi
so i divided both sides by 2 and got sin²x=1/2
so then I took the square root of both sides and got sinx= root(1/2)
I'm not quite sure what to do now? I'm not sure if i need a unit circle?

 

[Yogi]

2sin²x=1 where 0 is less than or equal to x and x is less than 2 pi
so i divided both sides by 2 and got sin²x=1/2
so then I took the square root of both sides and got sinx= root(1/2)
I'm not quite sure what to do now? I'm not sure if i need a unit circle?

Also,
\(\displaystyle \sin(x)=-\sqrt{\frac{1}{2}}\) (square both sides of this and get the original problem)
At this point, yes you could use a unit circle to find the solutions. It might look more familiar if you rationalize the denominator first.

 

[CrystalTennis]

oh okay so it becomes sinx= (root 2)/2
I forgot to rationalize the denominator, this makes so much more sense now thank you!

 

[Deleted member 4993]

You should have four (4) solutions in the given domain.