# Trigonometry General Solution

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?

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

## 3 Answers

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.

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

Knowing the following equality is always helpful for you:

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

• Indeed, forgot all about that identity, it's what I ended up using. Thanks! – Juxhin Oct 25 '14 at 11:27
• $\cos(2\theta) = \cos^2 - \sin^2\theta = 2\cos^2\theta - 1 = 1 - {\color{red}2}\sin^2\theta$. – N. F. Taussig Oct 25 '14 at 11:36
• @N.F.Taussig Yes, it is a type. – 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. $\cos\theta=0\lor\sin2\theta=1$.