Find the value of $\alpha $ given $2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha )$ Given:
$$2\sin\theta -\sqrt 5 \cos \theta  \equiv  - 3\cos (\theta  + \alpha ),$$
where $$0 <\alpha  < 90^\circ, $$
find $α.$
The issue I have with this question is the $-3$ on the right hand side, it really complicates things and I don't know how to deal with it.
When I usually come across equations of the form $$a\sin \theta  + b\cos \theta $$ it's a relatively straight forward process of converting them into another equation of the form
$$\sqrt {{a^2} + {b^2}} \left({a \over {\sqrt {{a^2} + {b^2}} }}\sin \theta  - {b \over {\sqrt {{a^2} + {b^2}} }}\cos \theta \right)$$
where $$\cos \alpha  = {a \over {\sqrt {{a^2} + {b^2}} }}$$
and $$\sin \alpha  = {b \over {\sqrt {{a^2} + {b^2}} }}$$
(or some other variant of this).
I cant wrap my head around this, specifically. If
$$2\sin\theta -\sqrt 5 \cos \theta  \equiv  - 3\cos (\theta  + \alpha ),$$
this must mean the square root expression prefixing the cosine addition identity must be $-3$, as
$$\sqrt {{a^2} + {b^2}} \cos(\theta  + \alpha ) \equiv  - 3\cos(\theta  + \alpha ).$$
I am aware the square root operation outputs negative values I just dont know how to deal with that in this instance.
I hope this doesn't read too convoluted, thank you.
 A: Your assumption is that the relation holds for all $\theta$. If that's really the assumption, then you could let $\theta$ be any angle, say $0$. 
Then $$\frac{\sqrt{5}}{3}=\cos(\alpha)$$ and $$\alpha=\arccos\left(\frac{\sqrt{5}}{3}\right)$$
A: Let  $2=r\sin\beta,\sqrt5=r\cos\beta$ where $r\ge0$
Squaring and adding we get $r^2=2^2+5=9\implies r=3$
and $\sin\beta=\frac23,\cos\beta=\frac{\sqrt5}3\implies \beta=\arccos \frac{\sqrt5}3=\arcsin \frac23$
So,$$2\sin\theta-\sqrt5\cos\theta=r\sin\beta\sin\theta-r\cos\beta\cos\theta=-r\cos(\theta+\beta)=-3\cos(\theta+\beta)$$
$$\implies 3\cos(\theta+\beta)=3\cos(\theta+\alpha)$$
$$\implies \cos(\theta+\beta)=\cos(\theta+\alpha)$$
So, $$\theta+\beta=2n\pi\pm (\theta+\alpha)$$
Taking the '+' sign, $\theta+\beta=2n\pi+ (\theta+\alpha)$
$\implies \alpha=\beta-2n\pi\equiv \beta\pmod{2\pi}$
Taking the '-' sign, $\theta+\beta=2n\pi- (\theta+\alpha)$
$\implies \alpha=-\beta+2n\pi-2\theta\equiv-(\beta+2\theta)\pmod{2\pi}$
A: $$\Rightarrow {\rm{2sin}}\theta {\rm{ - }}\sqrt 5 \cos \theta  =  - 3\cos (\theta  + \alpha )$$
$$\Rightarrow -\frac{2}{3}\sin \theta+\frac{\sqrt 5 }{3}\cos \theta=  \cos (\theta  + \alpha ) $$
$$\Rightarrow -\frac{2}{3}\sin \theta+\frac{\sqrt 5 }{3}\cos \theta=  \cos \theta \cos \alpha  - \sin \theta \sin \alpha$$
$$\Rightarrow \frac{\sqrt 5 }{3}\cos \theta -\frac{2}{3}\sin \theta = \cos \theta \cos \alpha  - \sin \theta \sin \alpha$$
Equating both sides and considering $0<\alpha < 90^0$ i.e. both $\cos \alpha$ and $\sin \alpha$ lie on the first quadrant.
$$\cos \alpha = \frac{\sqrt 5 }{3}\tag1$$
$$\sin \alpha = \frac{2}{3}\tag2$$
Dividing $(2)$ by $(1)$
$$\frac{\sin \alpha}{\cos \alpha} =\frac{\frac{2}{3}}{\frac{\sqrt 5 }{3}}=\frac{2}{sqrt(5)}$$
$$\tan \alpha = \frac{2}{\sqrt5}$$
$$\alpha = tan^{-1}\frac{2}{\sqrt5}$$

Note, the reason you went wrong because of your formulation, 
Note
$$\sqrt {{a^2} + {b^2}} \left({a \over {\sqrt {{a^2} + {b^2}} }}\sin \theta  - {b \over {\sqrt {{a^2} + {b^2}} }}\cos \theta \right) \ne \sqrt {{a^2} + {b^2}} \cos(\theta  + \alpha )$$
But rather
$$-\sqrt {{a^2} + {b^2}} \left({b \over {\sqrt {{a^2} + {b^2}} }}\cos - {a \over {\sqrt {{a^2} + {b^2}} }}\sin \theta   \theta \right) = - \sqrt {{a^2} + {b^2}} \cos(\theta  + \alpha )$$
Which will derive to
$$-\sqrt {{a^2} + {b^2}} \cos(\theta  + \alpha ) =  - 3\cos(\theta  + \alpha ).$$
A: You can use the so-called double angle formula. Notice that
$$-3\cos(\theta+\alpha) \equiv -3\cos\theta\cos\alpha + 3 \sin\theta\sin\alpha $$
Now you can compare coefficients. You want $-3\cos(\theta+\alpha) \equiv 2\sin\theta - \sqrt{5}\cos\theta$, and so you need to find an $\alpha$ for which $3\sin\alpha = 2$ and $3\cos\alpha = \sqrt{5}$. It follows that
$$\frac{3\sin\alpha}{3\cos\alpha} \equiv \tan\alpha = \frac{2}{\sqrt{5}}$$
You $\alpha$ is then $\arctan(2/\sqrt{5}) \approx 41.8^{\circ}$.
