# Trig solve for x [ Attempt Included ]

• lovemake1
In summary, we are solving the equation 2Cos^2(2x) + 1 = 3Cos(2x) for x in the interval [0,2pi). Using the substitution u=cos(2x), we get the quadratic equation 2u^2 - 3u + 1 = 0. Solving for u, we get u=1/2 and u=1. Substituting back in for cos(2x), we get two sets of solutions: x=pi/6, 5pi/6 and x=0, pi/2. It is important to remember that the function is periodic, so there will be multiple solutions.
lovemake1

## Homework Statement

solve the following equation for x in the interval [0,2pi)
2Cos^2(2x) + 1 = 3Cos(2x)

## Homework Equations

Cos(2x) = 2Cos^2 (x) - 1

## The Attempt at a Solution

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

2Cos(2x)Cos(2x) + 1 = 3Cos(2x)
2Cos(2x) + 1 = 3
2(2Cos^2 (x) - 1) - 2 =0
4Cos^2 (x) - 2 - 2 = 0
4cos^2 (x) = 4
Cos^2 (x) = 1
Cosx = 1

x = 0 is this right?
please correct me if i messed up thanks,

Last edited:
That's one answer, but there are more. Try the substitution u=cos(2x). That will give you a quadratic, which will give you both of the answers you need.

Last edited:
2cos^2(2x) + 1 = 3 cos(2x)
2cos^2(2x) - 3cos(2x) + 1 = 0

let u = cos(2x)

2u^2 - 3u + 1 = 0
(2u - 1 )(u - 1) = 0

u = 1/2, 1

Case 1. Cos2x = 1/2
cos2x = 1/2
2cos^2x - 1 = 1/2
2cos^2x = 1/2 - 1
2cos^2x - = -1/2
cos^2x = -1/4
cosx = sqrt(-1/4)

but how can you sqrt negative? is this error ?

Case 2. Cos2x = 1
2Cos^2x - 1 = 1
2Cos^2x - 2 =0
2(Cos^2x - 1) = 0
Cos^2x = 1
Cosx = 1

Did i get it right ?

Last edited:
Case 2 is correct, but in case 1, you should have added 1 instead of subtracting it.

Case 1:
cos(2x)=1/2
2cos^2(2x)-1=1/2
2cos^2(2x)=1/2+1=3/2
Try it from here.

And you have to remember that the function is periodic, so there will be multiple solutions to each of those.

Last edited:

## 1. How do I solve for x in a trigonometric equation?

Solving for x in a trigonometric equation involves using trigonometric identities and algebraic manipulation to isolate x on one side of the equation. The steps for solving may vary depending on the type of trigonometric equation given.

## 2. What are the basic trigonometric identities used for solving equations?

The basic trigonometric identities used for solving equations include the Pythagorean identities, the reciprocal identities, the quotient identities, and the cofunction identities. These identities can be used to simplify trigonometric expressions and equations.

## 3. How do I determine the domain of a trigonometric equation?

The domain of a trigonometric equation is the set of all possible values for the independent variable (usually x). To determine the domain, you must consider any restrictions on the trigonometric functions in the equation, such as division by zero or negative values under the square root.

## 4. Can I use a calculator to solve trigonometric equations?

Yes, a calculator can be used to solve trigonometric equations, especially for more complex equations involving multiple trigonometric functions. However, it is important to understand the steps and concepts behind solving the equation manually before relying solely on a calculator.

## 5. What are some common mistakes to avoid when solving trigonometric equations?

Some common mistakes to avoid when solving trigonometric equations include forgetting to use the correct trigonometric identity, not simplifying the equation enough before solving, and making calculation errors. It is also important to check for extraneous solutions, which may arise when manipulating equations involving trigonometric functions.

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