# Question #ac026

Sep 26, 2016

$\sin \left(2 \theta + \frac{\pi}{3}\right) = \frac{1}{2}$

$\implies \sin \left(2 \theta + \frac{\pi}{3}\right) = \sin \left(\frac{\pi}{6}\right)$

$\implies 2 \theta + \frac{\pi}{3} = \frac{\pi}{6}$

$\implies 2 \theta = - \frac{\pi}{6}$
$\implies \theta = - \frac{\pi}{12} \text{ not in the given range}$

Again

$\sin \left(2 \theta + \frac{\pi}{3}\right) = \frac{1}{2} = \sin \left(\pi - \frac{\pi}{6}\right)$

$\implies 2 \theta + \frac{\pi}{3} = \frac{5 \pi}{6}$

$\implies 2 \theta = \frac{5 \pi}{6} - \frac{\pi}{3}$

$\implies 2 \theta = \frac{3 \pi}{6} = \frac{\pi}{2}$,

$\implies \theta = \frac{\pi}{4}$

Again

$\sin \left(2 \theta + \frac{\pi}{3}\right) = \frac{1}{2} = \sin \left(2 \pi + \frac{\pi}{6}\right)$

$\implies 2 \theta + \frac{\pi}{3} = \frac{13 \pi}{6}$

$\implies 2 \theta = \frac{13 \pi}{6} - \frac{2 \pi}{6}$

$\implies 2 \theta = \frac{11 \pi}{6}$

$\implies \theta = \frac{11 \pi}{12}$

So for $0 \le \theta \le \pi$ the solutions are

$\theta = \frac{\pi}{4} \mathmr{and} \theta = \frac{11 \pi}{12}$

Sep 26, 2016

Due to the request for clarification in the question, this answer is a bit long. Feel free to skip over any unneeded explanations.

Unless otherwise specified, we generally consider the argument of a trigonometric function to be in radians. Radians are, in a sense, the natural way of measuring an angle within a circle, as they relate the radius of the circle to the angle measured (one radian is the measure of an angle subtended by an arc whose length equals the radius). A full circle is $2 \pi$ radians (equivalent to ${360}^{\circ}$), and as such, $\pi$ appears quite commonly in well known angles.

With that in mind, let's see what this problem is asking.

The above is a graph of $y = \sin \left(2 x + \frac{\pi}{3}\right)$, with dashed lines showing $y = \frac{1}{2}$ and $x = \pi$. Note that between $x = 0$ and $x = \pi$, the $y = \frac{1}{2}$ line intersects our graph at two points. The $x$-coordinates of those points are our solutions.

To figure out what those points are, we will consult the unit circle .

On the unit circle, the $x$ and $y$ coordinates tell us the value of $\cos \left(\theta\right)$ and $\sin \left(\theta\right)$, respectively. For example, from the circle, we see that $\sin \left(\theta\right) = \frac{1}{2}$ at $\theta = \frac{\pi}{6}$ and $\theta = \frac{5 \pi}{6}$.

(The angles shown on the circle and the values of sine and cosine at those angles are good to have memorized. As a side note, they can be derived by using the sohcahtoa definition on the special right triangles with angles ${30}^{\circ} - {60}^{\circ} - {90}^{\circ}$ or ${45}^{\circ} - {45}^{\circ} - {90}^{\circ}$. Additionally, because it is a circle, we can add or subtract any multiple of $2 \pi$ (${360}^{\circ}$) to an angle without changing the values of the trig functions.)

From the above, we now know that for $\sin \left(x\right)$ to equal $\frac{1}{2}$, we need $x = \frac{\pi}{6} + 2 \pi k$ or $x = \frac{5 \pi}{6} + 2 \pi k$ where $k$ is an integer. We can now solve the problem algebraically. Substituting our argument $2 \theta + \frac{\pi}{3}$ for $x$, we get

$2 \theta + \frac{\pi}{3} = \frac{\pi}{6} + 2 \pi k$

$\implies 2 \theta = - \frac{\pi}{6} + 2 \pi k$

$\implies \theta = - \frac{\pi}{12} + \pi k$

OR

$2 \theta + \frac{\pi}{3} = \frac{5 \pi}{6} + 2 \pi k$

$\implies 2 \theta = \frac{\pi}{2} + 2 \pi k$

$\implies \theta = \frac{\pi}{4} + \pi k$

Now we consider the restriction $0 \le \theta \le \pi$ to see what integers we can use for $k$. To get $\theta$ to stay within that range, we must have $k = 1$ for the first possibility, and $k = 0$ for the second. Thus, we get two answers:

$\theta = - \frac{\pi}{12} + 1 \pi = \frac{11 \pi}{12}$
OR
$\theta = \frac{\pi}{4} + 0 \pi = \frac{\pi}{4}$

Thus, we have the solution set $\theta \in \left\{\frac{\pi}{4} , \frac{11 \pi}{12}\right\}$.

Checking, we find that the equality hold for each of the above values:

$\sin \left(2 \left(\frac{\pi}{4}\right) + \frac{\pi}{3}\right) = \sin \left(\frac{\pi}{2} + \frac{\pi}{3}\right) = \sin \left(\frac{5 \pi}{6}\right) = \frac{1}{2}$

$\sin \left(2 \left(\frac{11 \pi}{12}\right) + \frac{\pi}{3}\right) = \sin \left(\frac{11 \pi}{6} + \frac{\pi}{3}\right) = \sin \left(\frac{\pi}{6} + 2 \pi\right) = \frac{1}{2}$