Solve Trig Inequality: 2cos^2(x) + 1 = 3cos(2x) [0,2pi)

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In summary, to solve for x in the interval [0,2pi), use the identity Cos(2x) = 2Cos^2(x) - 1 to rewrite 3cos(2x) as 3-6sin^2(x). Then, continue solving for x by simplifying the equation and using the quadratic formula if necessary.
  • #1
lovemake1
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Homework Statement


solve the following equations or inequalties for x in the interval [0,2pi)

2cos^2(x) + 1 = 3cos(2x)

Homework Equations





The Attempt at a Solution



My attempt at the problem:

2cosx(cos2x) + 1 = 3cos(2x)
2cosx(cos^2x - sin^2x) + 1 = 3cos(2x)
2cos^3x - 2cosxsin^2x + 1 = 3(1-2sin^2x)
2cos^3x - 2cosxsin^2x + 1 = 3-6sin^2x
2cos^3x - 2cosxsin^2x + 1 - 3 + 6sin^2x = 0

and I am getting lost here...


please help,
 
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  • #2
use this identity:
Cos(2x) = 2Cos^2(x) - 1
 
  • #3
lovemake1 said:

Homework Statement


solve the following equations or inequalties for x in the interval [0,2pi)

2cos^2(x) + 1 = 3cos(2x)


My attempt at the problem:
You have a mistake in your first step, below. 2cos2(x) is not equal to 2cos(x)*cos(2x). Use the identity that JonF gave to rewrite 3cos(2x) in terms of cos2(x).
lovemake1 said:
2cosx(cos2x) + 1 = 3cos(2x)
2cosx(cos^2x - sin^2x) + 1 = 3cos(2x)
2cos^3x - 2cosxsin^2x + 1 = 3(1-2sin^2x)
2cos^3x - 2cosxsin^2x + 1 = 3-6sin^2x
2cos^3x - 2cosxsin^2x + 1 - 3 + 6sin^2x = 0

and I am getting lost here...


please help,
 

1. How do I solve for x in this trig inequality?

To solve this trig inequality, we first need to use algebraic manipulation to get all the trigonometric functions on one side of the equation and all constants on the other side. In this case, we can subtract 1 from both sides and then use the double angle formula for cosine to simplify the right side of the equation. This will give us a quadratic equation that we can solve for the values of x.

2. What is the domain for x in this inequality?

The domain for x in this inequality is [0, 2π], which represents the interval of possible solutions for x.

3. Can I use a calculator to solve this inequality?

Yes, you can use a calculator to solve this inequality. However, it is important to remember that calculators may only give you approximate solutions, so it is always best to double-check your answers by hand.

4. What is the significance of the interval [0, 2π] in this inequality?

The interval [0, 2π] represents the full period of the trigonometric functions involved in this inequality. This means that any solutions in this interval will satisfy the inequality, but solutions outside of this interval will not.

5. Are there any special cases or restrictions for solving this inequality?

Yes, there are a few special cases and restrictions to keep in mind when solving this trig inequality. For example, since we are dealing with cosine functions, we must ensure that the values of x we find satisfy the restrictions for cosine, such as the range of -1 to 1. Additionally, we must also be aware of any extraneous solutions that may arise from our algebraic manipulations.

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