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

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In summary, the solutions for x in the equation cos(2x)=1/2 in the interval (0, 2∏) are x=∏/6, 5∏/6, and 13∏/6. The last solution can be found by considering the period of cosine and the range of 2x in the given interval.
  • #1
darshanpatel
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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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  • #2
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?
 
  • #3
What do you mean by corresponding range of 2x? I don't understand...
 
  • #4
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
 

1. How do you solve equations involving cos(2x) in the (0, 2∏) interval?

To solve an equation like cos(2x) = 1/2 in the interval (0, 2∏), you can use the double angle formula for cosine: cos(2x) = cos²(x) - sin²(x). Rewrite the equation as cos²(x) - sin²(x) = 1/2. Then, use the Pythagorean identity sin²(x) + cos²(x) = 1 to simplify the equation to 2cos²(x) - 1 = 1/2. Solving for cos(x), we get cos(x) = ± √(3)/2. Since we are looking for solutions in the interval (0, 2∏), we need to consider both the positive and negative solutions, which are cos(x) = √(3)/2 and cos(x) = -√(3)/2. This gives us two solutions for x: x = π/6 and x = 5π/6.

2. Can you use the unit circle to solve equations involving cos(2x)?

Yes, the unit circle can be a helpful tool in solving equations involving cos(2x). By plotting the cosine values on the unit circle, you can easily find the reference angles and use them to find the solutions in the given interval. In the example of cos(2x) = 1/2, the reference angle is π/6, which corresponds to the solutions x = π/6 and x = 5π/6.

3. Are there any other ways to solve equations involving cos(2x) in the (0, 2∏) interval?

Yes, there are other methods you can use to solve these types of equations. You can also use the half angle formula for cosine or the inverse cosine function. Additionally, you can graph both sides of the equation and find the intersection points, which represent the solutions.

4. How do you handle equations where cos(2x) is equal to a negative value?

If cos(2x) is equal to a negative value, such as -1/2 or -√(3)/2, you can follow the same steps as in question 1 to solve for x. However, keep in mind that the solutions you find will be in the form of an angle, so you may need to use inverse trigonometric functions to find the actual values for x.

5. Can you use the same method to solve equations involving other trigonometric functions?

Yes, you can use similar methods to solve equations involving other trigonometric functions such as sine and tangent. However, the identities and formulas used will be different. For example, to solve an equation like sin(2x) = 1/2, you would use the double angle formula for sine: sin(2x) = 2sin(x)cos(x). Then, use the Pythagorean identity and solve for sin(x), just like in question 1.

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