solve $\sin 2x + \sin x = 0$ using addition formula

$\sin 2x + \sin x = 0$

Using the addition formula, I know that

$\sin 2x = 2\sin x \cos x$

=> $2\sin x \cos x + \sin x = 0$

=> $\sin x(2\cos x + 1) = 0$

=> $\sin x = 0$ and $\cos x = -\frac{1}2$

I know that $\sin x = 0$ in first and second quadrant so $x = 0$ and $x = 180$

What I do not know is what to do with $\cos x = -\frac{1}2$ and which quadrants this applies to.

The book I got the question from gives the following answer which does not make sense to me:

0, 120, 180, 240, 360

• $\cos x<0$ for 2nd and 3rd quadrants, so $\cos x=-\frac{1}{2}$ for $180^o\pm60^o$ The $360^o$ is the same as $0^o$. – almagest Apr 12 '16 at 20:54
• It should say "or", not "and", i.e. $\sin x=0\text{ or }\cos x = -\frac 1 2$. $\qquad$ – Michael Hardy Apr 12 '16 at 21:03
• I never know how many values there are for x in these equations. where are the values 120, 240 and 360 coming from? – dagda1 Apr 12 '16 at 21:10
• I've accounted for every value for x apart from 360. why is 360 there? – dagda1 Apr 13 '16 at 19:48

We know that $\cos 60^\circ = \frac{1}{2}$.

The cosine of an angle is defined to be the $x$-coordinate of the point where the terminal side of an angle in standard position (initial side on the positive $x$-axis and vertex at the origin) intersects the unit circle. Therefore, the cosine function is positive if the terminal side of the angle lies in the first quadrant, fourth quadrant, or on the positive $x$-axis; $0$ if the terminal side of the angle lies on the $y$-axis; and negative if the terminal side of the angle lies in the second quadrant, third quadrant, or on the negative $x$-axis.

Now consider the diagram below. Observe that $\cos(\pi - \theta) = \cos(180^\circ - \theta) = -\cos\theta$. Hence, $$\cos(180^\circ - 60^\circ) = \cos(120^\circ) = -\cos(60^\circ) = -\frac{1}{2}$$

Also, observe that $\cos(\pi + \theta) = \cos(180^\circ - \theta) = -\cos\theta)$. Hence, $$\cos(180^\circ + 60^\circ) = \cos(240^\circ) = -\cos(60^\circ) = -\frac{1}{2}$$

$$\sin(2x)+\sin(x)=0$$

First, you have to remember that: $$\sin(a)+\sin(b)=2\sin(\frac{a+b}{2})\cos(\frac{a-b}{2})$$

Aplying this in the equation:

$$2\sin(\frac{3x}{2})\cos(\frac{x}{2})=0$$

$$\sin(\frac{3x}{2})\cos(\frac{x}{2})=0$$

Finally, you have two equations: $$\sin(\frac{3x}{2})=0$$ $$\cos(\frac{x}{2})=0$$

For the first, the solution are $x=\frac{2n\pi}{3} / n=0,1,2$

And for the second are $x=n\pi / n=1,3$

HINT

Cosine is the horizontal coordinate, so $\cos x < 0$ in II and III quadrants. Note that $$\cos(180^\circ - \theta) = -\cos \theta = \cos (180^\circ + \theta)$$

• Did you mean $\cos(180^\circ - \theta) = \color{red}{-}\cos\theta = \cos(180^\circ + \theta)$? – N. F. Taussig Apr 12 '16 at 21:13
• @N.F.Taussig fixed – gt6989b Apr 13 '16 at 0:00

If $\sin x=0$, then $x=k\pi$, where $k\in\mathbb{Z}$. If $\cos x=-\frac{1}{2}$, then either $x=\frac{2\pi}{3}+2k\pi, k\in\mathbb{Z}$ or $\cos x=\frac{4\pi}{3}+k\pi, k\in\mathbb{Z}.$

$\cos(x) = -1/2$ in the quadrants where $cos(x)$ is negative, so quadrants 2 and 3. You are correct about $\sin(x)$. Look at $\cos^{-1}(x)$ and see where it is $1/2$.

• I'm still confused about what the value of $\cos x = -\frac{1}2$ in these quadrants or where 120. 240 and 360 came from. $\cos x = \frac{1}2$ is root 3 over 2 but the equation had a negative value so I'm a bit lost – dagda1 Apr 12 '16 at 21:15