jonnburton
Junior Member
- Joined
- Dec 16, 2012
- Messages
- 155
I've been having more problems with trig equations involving identities. The following is a question I have been trying to solve without success. I'd be very grateful if anyone could give me any suggestions on this:
Solve the equation \(\displaystyle tan 2\theta = 3 tan\theta \), for \(\displaystyle 0 \le\theta \le 2\pi\)
The first thing I thought was to get rid of the \(\displaystyle 2\theta\)
\(\displaystyle \frac{2tan\theta}{1-tan^2\theta} = 3 than \theta\)
\(\displaystyle 2 tan\theta = 3 tan\theta (1 - tan^2\theta)\)
\(\displaystyle 2 tan \theta = 3 tan \theta - 3 tan^2\theta\)
\(\displaystyle 3 tan^3 \theta - tan \theta = 0\)
\(\displaystyle 3 tan^3 \theta = tan \theta\)
I thought it may make things more striaghtforward to put it in terms of sin and cos:
\(\displaystyle 3(\frac{sin^3 \theta}{cos^3 \theta}) = \frac{sin\theta}{cos\theta}\)
Multiplying all terms by \(\displaystyle cos^3\theta\)
\(\displaystyle sin\theta cos^2\theta = 3sin^3\theta\)
Divide by \(\displaystyle sin \theta\)
\(\displaystyle cos^2 \theta = 3 sin^2 \theta\)
Putting in terms of sin theta:
\(\displaystyle (1-sin^2 \theta) = 3 sin \theta\)
\(\displaystyle 1 = 4 sin^2 \theta\)
\(\displaystyle sin^2 \theta = \frac{1}{4}\)
\(\displaystyle sin \theta =\pm \sqrt\frac{1}{4}\)
Which would mean \(\displaystyle \theta = 0.253\) radians (or \(\displaystyle \theta = -0.253\) radians, which is outside the domain) and \(\displaystyle \theta = \pi - 0.253 = 2.89\) radians.
However, again my answer is wrong; according to the book, the solution includes 0, 5,24 and 2.62 radians (and several more values, up to and including 2 pi)
Solve the equation \(\displaystyle tan 2\theta = 3 tan\theta \), for \(\displaystyle 0 \le\theta \le 2\pi\)
The first thing I thought was to get rid of the \(\displaystyle 2\theta\)
\(\displaystyle \frac{2tan\theta}{1-tan^2\theta} = 3 than \theta\)
\(\displaystyle 2 tan\theta = 3 tan\theta (1 - tan^2\theta)\)
\(\displaystyle 2 tan \theta = 3 tan \theta - 3 tan^2\theta\)
\(\displaystyle 3 tan^3 \theta - tan \theta = 0\)
\(\displaystyle 3 tan^3 \theta = tan \theta\)
I thought it may make things more striaghtforward to put it in terms of sin and cos:
\(\displaystyle 3(\frac{sin^3 \theta}{cos^3 \theta}) = \frac{sin\theta}{cos\theta}\)
Multiplying all terms by \(\displaystyle cos^3\theta\)
\(\displaystyle sin\theta cos^2\theta = 3sin^3\theta\)
Divide by \(\displaystyle sin \theta\)
\(\displaystyle cos^2 \theta = 3 sin^2 \theta\)
Putting in terms of sin theta:
\(\displaystyle (1-sin^2 \theta) = 3 sin \theta\)
\(\displaystyle 1 = 4 sin^2 \theta\)
\(\displaystyle sin^2 \theta = \frac{1}{4}\)
\(\displaystyle sin \theta =\pm \sqrt\frac{1}{4}\)
Which would mean \(\displaystyle \theta = 0.253\) radians (or \(\displaystyle \theta = -0.253\) radians, which is outside the domain) and \(\displaystyle \theta = \pi - 0.253 = 2.89\) radians.
However, again my answer is wrong; according to the book, the solution includes 0, 5,24 and 2.62 radians (and several more values, up to and including 2 pi)
