Analytic Trigonometry: Solving Equations in (0, 2∏) Interval for cos(2x)=1/2

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Homework Statement



Find all the solutions of each equation in the interval (0, 2∏)

cos(2x)=1/2

Homework Equations



-None-

The Attempt at a Solution



cos(2x)=1/2
2x=(cos^(-1))*1/2
2x= ((∏)/3), ((5∏)/3)
2x= ((∏)/3), ((5∏)/3), ((14∏)/6)
x= ∏/6, ((5∏)/6),

That's all I have.

I know there is one more solution, but I don't know how to find it.

Please help, and thanks in advance.
 
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darshanpatel said:

Homework Statement



Find all the solutions of each equation in the interval (0, 2∏)

cos(2x)=1/2


Homework Equations



-None-

The Attempt at a Solution



cos(2x)=1/2
2x=(cos^(-1))*1/2
2x= ((∏)/3), ((5∏)/3)
2x= ((∏)/3), ((5∏)/3), ((14∏)/6)
x= ∏/6, ((5∏)/6),

That's all I have.

I know there is one more solution, but I don't know how to find it.

Please help, and thanks in advance.

Sure there's just the one more?

You're asked to find x in (0,2∏). What's the corresponding range of 2x?

You know that cosine has period 2∏. Can you use this to figure out what other values 2x can take, and the corresponding x values?
 
What do you mean by corresponding range of 2x? I don't understand...
 
darshanpatel said:
What do you mean by corresponding range of 2x? I don't understand...

If min(x) = 0, max(x) = 2pi, what are the corresponding min and max values 2x can take? Start with that. That way you'll know what values of 2x to consider when you take the inverse cosine of 0.5