I've been working out some general equations recently, whilst the simple ones are fairly easy to work with, the more advanced ones (for me ofcourse) seem to somewhat confuse me since I have no background on this topic.

I am simply asked to give the General Solution for the given equations. A couple I've had issues with are the following:

  • $\cos(2\theta) = \sin(\theta)$
  • $2\sin^2(\theta) - \sin(\theta) = 1$
  • $\cos(\theta) \sin(2\theta) = \cos(\theta)$

To get an idea of how I worked them - $$\sin(\theta) = \cos(\frac{\pi}{2} - \theta )$$

$$\therefore \cos(2\theta) = \cos(\frac{\pi}{2} - \theta )$$

Let $\theta$ = 2$\theta$ and $\alpha = \frac{\pi}{2} - \theta$ in the formula, $\theta = 2n\pi \pm \alpha $

$$ \therefore 2\theta = 2n\pi \pm \frac{\pi}{2} - \theta$$

From here on I simply solve the equation and leave theta as subject, which isn't all too relevant to the question. The approach is similar for all the questions with their respective formulae.

I presume I'm missing something crucial at this point and would like to ask, how do you solve the problems above?

  • $\begingroup$ Why have you tagged it Calculus? $\endgroup$ – Swapnil Rustagi Oct 25 '14 at 10:41
  • $\begingroup$ Excuse me, removed it and left Trigonometry only. $\endgroup$ – Juxhin Oct 25 '14 at 10:44

For $(1)$:

Use that $\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$ and that $\cos^2(\theta)=1-\sin^2(\theta)$.

Rewriting the equation gives: $$-2\sin^2(\theta)-\sin(\theta)+1=0$$

Substitute $$y=\sin(\theta)$$ and solve the quadratic equation.

The other questions are all rewritable as a quadratic equation.

  • $\begingroup$ Thanks, just what I needed! $\endgroup$ – Juxhin Oct 25 '14 at 10:52

Knowing the following equality is always helpful for you:

$\cos 2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta; \sin2\theta=2\sin\theta\cos\theta$

  • $\begingroup$ Indeed, forgot all about that identity, it's what I ended up using. Thanks! $\endgroup$ – Juxhin Oct 25 '14 at 11:27
  • $\begingroup$ $\cos(2\theta) = \cos^2 - \sin^2\theta = 2\cos^2\theta - 1 = 1 - {\color{red}2}\sin^2\theta$. $\endgroup$ – N. F. Taussig Oct 25 '14 at 11:36
  • $\begingroup$ @N.F.Taussig Yes, it is a type. $\endgroup$ – Paul Oct 25 '14 at 11:42
  1. $1-2s^2=s$. By inspection, $s=-1$ is a root and we can factor $2s^2+s-1=(2s-1)(s+1)$.
  2. $2s^2-s=1$. By inspection, $s=1$ is a root and we can factor $2s^2+s-1=(2s+1)(s-1)$.

  3. $\cos\theta=0\lor\sin2\theta=1$.


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