This may seem fairly straightforward, but I have been stuck on this for the past half-hour.

I need to use Double Angle Formulae such as the following:

  • $\sin2A ≡ 2\sin A \cos A$
  • $\cos2A ≡ \cos^2A - \sin^2A$
  • $\tan2A ≡ \frac{2\tan A}{1 - \tan^2A}$


  • $1 + \cos 2A ≡ 2\cos^2 A$
  • $1 - \cos 2A ≡ 2\sin^2A$

to solve this equation for all values of $\theta$ between $0^o < \theta <360^o$:

  • $3\tan\theta = 2\cos\theta$

I understand all of the identities above and how you get there, and I understand how to find the other values of $\theta$ between $0^o$ and $360^o$ once I have found one. I just get stuck solving this equation. Any help would be greatly appreciated.



I'm not really sure why you need to use double angle formula (some exercise?), but I'd just do it in most straightforward way \begin{align} 3 \tan \theta &= 2 \cos \theta \\ 3 \sin \theta &= 2 \cos^2 \theta = 2 - 2\sin^2 \theta \end{align} $$ 2\sin^2 \theta + 3 \sin \theta - 2 = 0 \\ \sin \theta = \frac {-3 \pm \sqrt{9+16}}{4} = \frac {-3 \pm 5}{4} = \left [ \begin{array}{l} -2 \text{ (spurious solution)}\\ \frac 12 \text{ (correct solution)} \end{array}\right . $$ So, final answer is $\theta \in \left \{ \frac \pi 6, \frac {5 \pi}6\right \}$ or $\theta \in \left \{ 30^\circ, 150^\circ\right \}$ if operating with degrees.

  • $\begingroup$ Thanks. Yes it was an exercise that required me to do it using double angle forumla $\endgroup$ – arch Oct 23 '13 at 17:42

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