1. ## should this factorise?

Start of problem at the top
I believe i have done it right but it doesnt seem to factorise
Do i need to run it from the quadratic equation?

2. Originally Posted by 200001

Start of problem at the top
I believe i have done it right but it doesnt seem to factorise
Do i need to run it from the quadratic equation?
$\displaystyle \sin{\theta} + \cos^2{\theta} = \frac{1}{5}$

$\displaystyle \sin{\theta} + 1 - \sin^2{\theta} = \frac{1}{5}$

$\displaystyle 0 = \sin^2{\theta} - \sin{\theta} - \frac{4}{5}$

Now Complete the Square:

$\displaystyle 0 = \sin^2{\theta} - \sin{\theta} + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 - \frac{4}{5}$

$\displaystyle 0 = \left(\sin{\theta} - \frac{1}{2}\right)^2 - \frac{1}{4} - \frac{4}{5}$

$\displaystyle 0 = \left(\sin{\theta} - \frac{1}{2}\right)^2 - \frac{21}{20}$

$\displaystyle \frac{21}{20} = \left(\sin{\theta} - \frac{1}{2}\right)^2$

$\displaystyle \pm\sqrt{\frac{21}{20}} = \sin{\theta} - \frac{1}{2}$

$\displaystyle \pm \frac{\sqrt{21}}{2\sqrt{5}} = \sin{\theta} - \frac{1}{2}$

$\displaystyle \pm \frac{\sqrt{105}}{10} = \sin{\theta} - \frac{1}{2}$

$\displaystyle \frac{1}{2} \pm \frac{\sqrt{105}}{10} = \sin{\theta}$

$\displaystyle \frac{5 \pm \sqrt{105}}{10} = \sin{\theta}$

$\displaystyle \theta = \arcsin{\left(\frac{5 + \sqrt{105}}{10}\right)}$ or $\displaystyle \theta = \arcsin{\left(\frac{5 - \sqrt{105}}{10}\right)}$.

3. Yes, thats where i went with it but this is from a book that isnt driven at any equations that wont factorise and the level its picthed at would not go into that depth.

4. Hello 200001
Originally Posted by 200001

Start of problem at the top
I believe i have done it right but it doesnt seem to factorise
Do i need to run it from the quadratic equation?
You have a sign wrong in the third line of your solution. It should be:
$\displaystyle \sin^2\phi-\sin\phi=\frac45$

$\displaystyle \Rightarrow 5\sin^2\phi-5\sin\phi-4=0$

$\displaystyle \Rightarrow \sin\phi = \frac{5\pm\sqrt{105}}{10}$
Now $\displaystyle -1\le\sin\phi\le 1$. So the only valid root is:
$\displaystyle \sin\phi = \frac{5-\sqrt{105}}{10}$

$\displaystyle \Rightarrow \phi = n\pi +(-1)^n \arcsin\left(\frac{5-\sqrt{105}}{10}\right)$