solve 2tanx cos2x+1=tan3x+2cos2x
2tan3x cos2x + 1 = tan3x + 2cos2x
Or ,2tan3x cos2x - tan3x - 2cos2x + 1 = 0
Or, tan3x ( 2cos2x - 1) - ( 2cos2x- 1) = 0
Or , (tan3x-1) (2cos2x-1) =0
Either, tan3x-1=0. Or, 2cos2x= 1
Or, tan3x = 1. Or,cos 2x= cos 1/2
Or, tan 3x=tan 45 Or, 2x=2nπ+_ π/3
Or, 3x=nπ+π/4 Or, x=nπ +_π/6
Or, x=nπ/3+π/12 •°• x=nπ +_π/6
•°• x=nπ/3+π/12
•°• x = nπ/3
[1-sin2xdx=
I apologize for any confusion, but the solution provided in the previous response is incorrect. Here is the correct solution:
Starting with 2tan3x cos2x + 1 = tan3x + 2cos2x, we can rearrange and factor to get:
2tan3x cos2x - tan3x - 2cos2x + 1 = 0
(tan3x - 1)(2cos2x - 1) = 0
So either tan3x = 1 or cos2x = 1/2.
For the first case, tan3x = 1 has solutions at 3x = π/4 + nπ, where n is any integer.
For the second case, we have:
cos2x = 1/2
2x = ±π/3 + 2nπ, where n is any integer.
So x = π/6 + nπ/2 or x = π/3 + nπ/2.
Therefore, the solutions are:
x = π/6 + nπ/2 (corresponding to cos2x = 1/2)
x = π/3 + nπ/2 (corresponding to cos2x = 1/2)
3x = π/4 + nπ (corresponding to tan3x = 1)
Oh, solving equations. That's always fun, isn't it? Let's dive right in!
2tan(x)cos(2x) + 1 = tan(3x) + 2cos(2x)
To make things a little less complicated, let's deal with the trigonometric functions one at a time. First, we have the term 2tan(x)cos(2x) on the left side.
Now, you know what they say about tangents... they're a bit touchy. So let's convert tan(x) into sine(x) over cosine(x).
2(sin(x) / cos(x)) * cos(2x) + 1 = tan(3x) + 2cos(2x)
Next up, we have the cosine term: cos(2x). Remember the double angle formula?
cos(2x) = 2cos^2(x) - 1
Let's substitute that into our equation:
2(sin(x) / cos(x)) * (2cos^2(x) - 1) + 1 = tan(3x) + 2(2cos^2(x) - 1)
Now, we simplify:
4sin(x)cos^2(x) - 2sin(x) + 1 = tan(3x) + 4cos^2(x) - 2
Can you still hang in there? Great!
Now, let's tackle the tangent on the right side. Remember the tangent sum formula?
tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)tan(b))
So, plug in the values:
tan(3x) = (tan(x) + tan(2x)) / (1 - tan(x)tan(2x))
We'll substitute that into our equation:
4sin(x)cos^2(x) - 2sin(x) + 1 = ((tan(x) + tan(2x)) / (1 - tan(x)tan(2x))) + 4cos^2(x) - 2
Phew! We're almost there, hang on!
Now, let's get rid of those tangents by converting them into sines and cosines:
4sin(x)cos^2(x) - 2sin(x) + 1 = ((sin(x) / cos(x)) + (sin(2x) / cos(2x))) / (1 - (sin(x) / cos(x))(sin(2x) / cos(2x))) + 4cos^2(x) - 2
Now, simplify, simplify, simplify! Combine like terms, cross-multiply, do whatever math magic you need to do.
Then, hopefully, you'll end up with a solution that puts a smile on your face (and my clown makeup on mine).
Keep in mind that solving trigonometric equations can sometimes get quite messy, so it might be a good idea to take a big sip of coffee or a deep breath before diving into it. Good luck, my friend!
To solve the equation 2tanx cos2x + 1 = tan3x + 2cos2x, let's break down the problem step by step:
1. Use trigonometric identities to simplify the equation:
Start by simplifying tan3x into terms of tanx:
tan3x = tan(2x + x)
Applying the tangent sum formula:
tan(2x + x) = (tan2x + tanx) / (1 - tan2x * tanx)
2. Substitute the trigonometric identity cos2x = 1 - sin^2(2x) into the equation:
2tanx(1 - sin^2(2x)) + 1 = (tan2x + tanx) / (1 - tan2x * tanx) + 2(1 - sin^2(2x))
3. Simplify the equation and collect like terms:
Expand both sides of the equation:
2tanx - 2tanx * sin^2(2x) + 1 = tan2x + tanx + 2 - 2sin^2(2x)
Rearrange the terms:
2 - 1 + 2sin^2(2x) = tan2x - 2tanx + tanx - 2tanx * sin^2(2x)
Combine like terms:
1 + 2sin^2(2x) = tan2x - 3tanx - 2tanx * sin^2(2x)
4. Use trigonometric identities to further simplify the equation:
Using the identity tan2x = 2tanx / (1 - tan^2(x)), substitute the expression for tan2x:
1 + 2sin^2(2x) = 2tanx / (1 - tan^2(x)) - 3tanx - 2tanx * sin^2(2x)
Multiply both sides by (1 - tan^2(x)) to eliminate the denominator:
(1 - tan^2(x))(1 + 2sin^2(2x)) + 3tanx(1 - tan^2(x)) + 2tanx * sin^2(2x) = 2tanx
Expand and collect like terms:
(1 - tan^2(x)) + 2sin^2(2x) - 2sin^2(2x)tan^2(x) + 3tanx - 3tanx * tan^2(x) + 2sin^2(2x)tanx = 2tanx
Combining terms:
1 - tan^2(x) + 3tanx - 3tanx * tan^2(x) = 2tanx
Rearrange the equation:
1 - tan^2(x) - 2tanx + 3tanx * tan^2(x) = 0
5. Solve the quadratic equation:
Let tanx = p to simplify the equation:
1 - p^2 - 2p + 3p * p^2 = 0
Simplify further:
1 - p^2 - 2p + 3p^3 = 0
Rearrange the equation:
3p^3 - p^2 - 2p + 1 = 0
Factoring or using the rational root theorem, we can find the solutions for p (tanx):
Upon solving the equation, the possible values of p (tanx) are p = -1/3, 1, or 1/3.
6. Solve for x:
Now, substitute p values back to tanx = p:
For p = -1/3:
tanx = -1/3
Using the inverse tangent function, we can find the angles:
x = arctan(-1/3) + kπ, where k is an integer.
Similarly, solve for x when p = 1 and p = 1/3.
Apply the arctan function to find the corresponding angles.
By following these steps, you can find the values of x that solve the given equation.
Good luck with that! I have no clue how to get an exact solution. The graphs at wolframalpha.com and the analysis provided there don't give me much hope:
http://www.wolframalpha.com/input/?i=2tanx+cos2x%2B1%3Dtan3x%2B2cos2x
2tan3x cos2x + 1 = tan3x + 2cos2x
Or ,2tan3x cos2x + 1 - tan3x - cos2x= 0
Or, tan3x ( cos2x - 1) - ( cos2x- 1) = 0
Or , (tan3x-1) (cos2x-1) =0
Either, tan3x-1=0. Or, cos2x= 1
Or, tan3x = 1. Or, 2x= (4n+1)π/4
Or, 3x = nπ •°• x= (4n+1)π/12
•°• x = nπ/3