sin2x=cos3x, solve for x

Results 1 to 5 of 5

- November 18th 2006, 03:26 PM #1

- Joined
- Nov 2006
- Posts
- 2

- November 18th 2006, 03:34 PM #2

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- November 18th 2006, 03:36 PM #3

- November 18th 2006, 04:02 PM #4

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 11,919
- Thanks
- 783

Hello, lexmex1546!

Here's a "cute" way to approach this problem . . .

Solve for

. . we see that and are angles in the*same right triangle*. . .

Code:* * * * 3x * * * * * * * * 2x * * * * * * * * *

Therefore: .

Of course, this is__not__a complete solution . . .

- November 18th 2006, 06:55 PM #5
The hard way.

Thus:

Pull everything to the LHS and factor the common cos(x):

So either or

For the sin solution, use the quadratic formula to find:

So

So the whole solution set in the domain will be x = from the solution and x = from the sine solution. These may be easily seen to be the same solutions as ThePerfectHacker gave you, only from a much messier derivation.

-Dan