Q:

$-1.4\sin 3x - 0.2 \cos 3x$ in the form $R \sin (3x+\alpha)$ such that $R>0$ and $0<\alpha<2\pi$

Write $-1.4 \sin 3x - 0.2 \cos 3x$ in the form $R \sin (3x+\alpha)$ such that $R>0$ and $0<\alpha<2\pi$

I found  $R= \sqrt{(-1.4)^2+(-0.2)^2}= \sqrt{2}$
And $\alpha= \arctan \frac{0.2}{1.4}= \arctan \frac {1}{7}$
Now the problem is  I could write this as 
$-\sqrt{2}\sin(3x+\arctan \frac{1}{7})$
But  this would violate the rule $R>0$ , I don't know how to write in the way they want.

A:

$$R \sin (3x+ \alpha)= R \sin 3x \cos \alpha + R \cos 3x \sin \alpha= -1.4 \sin 3x - 0.2 \cos 3x$$
$$R= \sqrt 2$$
$$\sqrt 2 \cos \alpha= -1.4$$
$$\sqrt 2 \sin \alpha =-0.2$$
$$\alpha= \pi + 0.14=3.28$$

