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 $\displaystyle \theta = \arcsin{\left(\frac{5 - \sqrt{105}}{10}\right)}$.
Hello 200001You have a sign wrong in the third line of your solution. It should be:$\displaystyle \sin^2\phi-\sin\phi=\frac45$Now $\displaystyle -1\le\sin\phi\le 1$. So the only valid root is:
$\displaystyle \Rightarrow 5\sin^2\phi-5\sin\phi-4=0$
$\displaystyle \Rightarrow \sin\phi = \frac{5\pm\sqrt{105}}{10}$
$\displaystyle \sin\phi = \frac{5-\sqrt{105}}{10}$Grandad
$\displaystyle \Rightarrow \phi = n\pi +(-1)^n \arcsin\left(\frac{5-\sqrt{105}}{10}\right)$