# General Solution of $\sin\theta=3\cos\theta$

I'm a high school level maths student currently working through some exercises for the general solution of trigonometric equations and have come across this one that I am stuck on. Any hints would be much appreciated!

Question:

Determine the general solution of the trigonometric equation: $\sin\theta=3\cos\theta$

So the first thing I think of doing when seeing this is making sure either side has the same ratio. I can do this using co-ratios, so:

$\sin\theta=3\sin(90^\circ-\theta)$
$\sin\theta=3\cos\theta$

From here I get stuck; there are no values to compute a reference angle with.

The textbook gives the solution as:

$\theta=71.6^\circ+k\cdot 180^\circ, k\in\mathbb{Z}$

Can anyone give me a hint? Thanks in advance!

-Shaun

• Read this to learn how to typeset mathematics in your question, then edit your question to use Mathjax. – Joel Reyes Noche Jul 16 '15 at 13:38
• Clue: $\frac{\sin(\theta)}{\cos(\theta)}=\tan(\theta)$ – Warren Hill Jul 16 '15 at 13:39

(I assume you know how to use the inverse trigonometric functions of a calculator.) Divide both sides by $\cos\theta$. What do you get?

• If i divide by Cosθ then I will get tanθ = 3(1) which gives an angle of 71.6° using a calculator. Hence θ = 71.6° + k.180° Thank you! – ShaunS Jul 16 '15 at 13:38
• You're welcome. Don't forget to indicate that the angle is measured in degrees. – Joel Reyes Noche Jul 16 '15 at 13:39
• Before dividing by Cosθ, be sure to check that there cannot be a solution such that Cosθ = 0 – P. Camilleri Jul 16 '15 at 14:13

Dividing both sides by $\cos \theta$ yields $$\frac{\sin \theta}{\cos \theta} = 3\frac{\cos \theta}{\cos \theta}$$ but we know that $\tan \theta = \frac{\sin \theta}{\cos \theta}$ so we get $$\tan \theta = 3$$

Hence $$\theta = \arctan 3 + n\pi, \quad \text{for some integer n}.$$

Or, if you want $$\theta = \arctan 3 + n\cdot 180^{\circ}, \quad \text{for some integer n}.$$

Notice, $$\sin\theta=3\cos\theta$$ $$\implies 3\cos\theta-\sin\theta=0$$ Now, divide the above equation by $\sqrt{3^2+(-1)^2}=\sqrt{10}$ $$\frac{3}{\sqrt{10}}\cos\theta-\frac{1}{\sqrt{10}}\sin\theta=0$$ Now, let $\frac{3}{\sqrt{10}}=\cos\alpha \implies \sin\alpha=\frac{1}{\sqrt{10}}$, we have $$\cos\theta\cos\alpha-\sin\theta\sin\alpha=0$$ $$\implies \cos(\theta+\alpha)=0$$ Now, writing the general solution, we get $$\theta+\alpha=(2n+1)\frac{\pi}{2}$$ $$\implies \theta=(2n+1)\frac{\pi}{2}-\alpha$$ Substituting the value of $\alpha$ $$\implies \color{blue}{\theta=(2n+1)\frac{\pi}{2}-\cos^{-1}\left(\frac{3}{\sqrt{10}}\right)}$$ $$\text{Or} \quad \color{blue}{\theta=(2n+1)\frac{\pi}{2}-\sin^{-1}\left(\frac{1}{\sqrt{10}}\right)}$$ Where, $\color{blue}{\text{n is any integer} }$