<s>[INST] solve $3\sin\theta+4\cos\theta=0$ Solve for $0 < \theta < 360$ 
Question 
$3 \sin \theta + 4 \cos \theta = 0$
Please help. I really can't figure this out 
Thanks :)
What I have tried
I tried using the a $\sin \theta + b \cos \theta = r \sin (\theta + a)$ rule but didn't work out (or I might have done it wrong)
 [/INST] Divide both sides by 5, and observe that the coefficients satisfy 
$\left(\frac{3}{5}\right)^2+ \left(\frac{4}{5}\right)^2=1.$ Consequently there is some angle $\alpha$ such that $\cos \alpha = 3/5, \sin\alpha =4/5$. (For comparison with other answers, note that $\tan\alpha = 4/3$). With this in mind we can use the sum-to-product identity to write
$$ \sin(\alpha+\theta) = \cos\alpha \sin \theta+\sin\alpha \cos\theta = \frac{3}{5}\sin\theta+\frac{4}{5}\cos\theta = 0.$$ So all that's left is to find the solutions of this equation.
</s>