If $a\sin x + b\cos(x+\theta) +b\cos(x-\theta) = d$, then what is the minimum value of $|\cos\theta|$? 
If $$a \sin x + b \cos(x+\theta)+ b \cos(x-\theta) = d$$
  then what is the minimum value of $|\cos\theta|$?
The answer is given: $\dfrac{\sqrt{d^2 - a^2}}{2|b|}$

I tried simplifying the equation to $a \sin x + 2b \cos\theta \cos x = d$, but I am stuck here. What must I do next?

$$d=a\sin x+2b\cos\theta\cos x=\sqrt{a^2+(2b\cos\theta)^2}\sin\left(x+\arcsin\dfrac{2b\cos\theta}{\sqrt{a^2+(2b\cos\theta)^2}}\right)$$
As for real $y,\sin y\le1$
$$\implies d\le\sqrt{a^2+(2b\cos\theta)^2}$$
Can you square both sides & rearrange?
