# Can anyone help me on this? Solve sin 2 theta + sin theta - tan theta = 0 for 0 < or = theta = or > 360

May 1, 2018

$\theta = {0}^{\circ} , {60}^{\circ} , {180}^{\circ} , {300}^{\circ}$ and ${360}^{\circ}$ if I'm making out the question correctly.

#### Explanation:

$\sin 2 \theta + \sin \theta - \tan \theta = 0$

$2 \sin \theta \cos \theta + \sin \theta - \sin \frac{\theta}{\cos} \theta = 0$

We can assume $\cos \theta \ne 0$. Let's multiply to clear the fraction.

$2 \sin \theta {\cos}^{2} \theta + \sin \theta \cos \theta - \sin \theta = 0$

$\sin \theta \left(2 {\cos}^{2} \theta + \cos \theta - 1\right) = 0$

$\sin \theta = 0$ or $2 {\cos}^{2} \theta + \cos \theta - 1 = 0$

$\theta = {0}^{\circ} \mathmr{and} {180}^{\circ} \mathmr{and} {360}^{\circ}$ from the first one.

$\left(2 \cos \theta - 1\right) \left(\cos \theta + 1\right) = 0$

$\cos \theta = \frac{1}{2} \mathmr{and} \cos \theta = - 1$

The first is trig's go-to triangle. $\theta = {60}^{\circ} \mathmr{and} {300}^{\circ}$

The second is $\theta = {180}^{\circ}$ which we already had.

Check:

theta =0 quad sin 0 + sin 0 - tan 0 = 0 quad sqrt

theta=60^circ quad sin 120+ sin 60-tan60=\sqrt{3}/2+\sqrt{3}/2- \sqrt{3} = 0 quad sqrt

theta=180^circ quad sin 360 + sin 180 - tan 180= 0+0-0=0 quad sqrt

$\theta = {300}^{\circ} \quad \sin 600 + \sin 300 - \tan 300$$= \sin \left(- 120\right) + \sin \left(- 60\right) - \tan \left(- 60\right)$= (-\sqrt{3}/2) + (-\sqrt{3}/2) - (-\sqrt{3}) = 0 quad sqrt