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For the trigonometric equation:


defined by $0 < \theta < 360$

I acquired the solutions of $75, 165, 255, 345, 435, 525$

Would anyone care to confirm if I am correct please?

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Look at my edited answer. – user159870 Aug 10 '14 at 12:49
up vote 1 down vote accepted

No. You need to be careful about $0^\circ\lt \theta \lt 360^\circ\ (\iff 0^\circ \lt 3\theta\lt 1080^\circ)$.

We have the form of $$3\theta =180^\circ+30^\circ+360^\circ \cdot k,\ 360^\circ -30^\circ +360^\circ \cdot k.$$ So, the answer is $$3\theta=210^\circ, 330^\circ, 570^\circ, 690^\circ, 930^\circ, 1050^\circ$$ $$\iff \theta=70^\circ, 110^\circ, 190^\circ, 230^\circ, 310^\circ, 350^\circ.$$

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Thanks. That was one of the other solutions I got. Still different from my answer sheet. I wish I can accept an answer in less than 20 minutes. – Qwerty Aug 10 '14 at 12:45

$$\sin 3 \theta= - \frac{1}{2}$$

$$\Rightarrow 3 \theta =2 \pi n - \frac{\pi }{6} \text{ or } 3 \theta =2 \pi n + \frac{7 \pi }{6}$$

$$\Rightarrow \theta=\frac{2 n \pi}{3}-\frac{\pi}{18} \text{ or } \theta=\frac{ 2 \pi n }{3}+\frac{7 \pi }{18}$$

$$\Rightarrow \theta=\frac{12 n \pi-\pi}{18} \text{ or } \theta=\frac{ 12 \pi n +7 \pi}{18}$$

Find for different $n \in \mathbb{N}_0$ the $\theta$ such that $0^{\circ}<\theta<360^{\circ}$.


  • $n=0$


  • $n=1$

$\theta=110^{\circ}$ or $\theta=190^{\circ}$

  • $n=2$

$\theta=230^{\circ}$ or $\theta=310^{\circ}$

  • $n=3$


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Thank you this method will be noted. – Qwerty Aug 10 '14 at 12:51

$\sin\alpha=\sin\beta\iff\alpha=\beta+k\times360^{0}$ or $\alpha=180^{0}-\beta+k\times360^{0}$.

This for integers $k$.

Apply that on $\alpha=3\theta$ and $\beta=150^{0}$ to find expressions in $k$ for $\theta$.

(Here $\sin\beta=-\frac{1}{2}$, so that $\sin\alpha=\sin\beta\iff2\sin3\theta=-1$)

Select the outcomes in $\left(0^{0},360^{0}\right)$.

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