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

Question

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.

Answer 1

$R= \sqrt 2$

$\sqrt 2 \cos \alpha= -1.4$

$\sqrt 2 \sin \alpha =-0.2$

Both $\sin \alpha$ and $\cos \alpha$ are negative in the 3rd quadrant

$\alpha= \pi + 0.14=3.28$

Answer 2

$$R \sin (3x+ \alpha)= R \sin 3x \cos \alpha + R \cos 3x \sin \alpha= -1.4 \sin 3x - 0.2 \cos 3x$$

$$R= \sqrt 2$$

$$\sqrt 2 \cos \alpha= -1.4$$

$$\sqrt 2 \sin \alpha =-0.2$$

$$\alpha= \pi + 0.14=3.28$$