# solve $3\sin\theta+4\cos\theta=0$

Solve for $0 < \theta < 360$

Question

$3 \sin \theta + 4 \cos \theta = 0$

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)

• Hint: convert to a ratio relationship. – Gina Jul 14 '14 at 1:46
• @Gina Why don't you make that an answer? (Maybe be a little more explicit.) – Cameron Williams Jul 14 '14 at 1:50
• Which do you use 360 (rad), or 360 (degree)? – choco_addicted Nov 29 '18 at 6:26

$3\sin \theta +4\cos\theta =0$

Notice that $\theta=\frac{\pi}{2}+\pi k$ is not a solution to this equation, and so it is valid to divide by $\cos \theta$:

$3\tan \theta =-4$, $\tan \theta =-\frac{4}{3}$

And so $\theta$ is....?

HINT: divide the ecuation by $\cos\theta$

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.