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Section 3.3 #11: Solve on the interval $0 \leq \theta \leq 2\pi$: $$2 \sin(\theta)+3=2.$$
Solution: First we isolate $\sin(\theta)$ on the left by subtracting by $3$ and then dividing by $2$: $$\sin(\theta) = -\dfrac{1}{2}.$$ Now we observe the unit circle to find places where $\sin(\theta) = -\dfrac{1}{2}$:

From this we see that the solution is $\theta = \dfrac{7\pi}{6}$ and $\theta=\dfrac{11\pi}{6}$.

Section 3.3 #16: Solve on the interval $0 \leq \theta \leq 2\pi$: $$4\cos^2(\theta)-3=0.$$
Solution: We must isolate $\cos(\theta)$. To do this first add $3$ to both sides and divide by $4$ to get $$\cos^2(\theta) = \dfrac{3}{4}.$$ Now take the square root of both sides to get $$\cos(\theta) = \pm \sqrt{ \dfrac{3}{4} } = \pm \dfrac{\sqrt{3}}{2}.$$ Now we observe the unit circle to find places where $\cos(\theta) = \dfrac{\sqrt{3}}{2}$ or $\cos(\theta) = -\dfrac{\sqrt{3}}{2}$:

From this we see that the solution is $\theta=\dfrac{\pi}{6}$, $\theta=\dfrac{5\pi}{6}$, $\theta=\dfrac{7\pi}{6}$, and $\theta=\dfrac{11\pi}{6}$.

Section 3.3 #19: Solve on the interval $0 \leq \theta \leq 2\pi$: $$\cos(2\theta)=-\dfrac{1}{2}.$$ Solution: We observe the unit circle to find the values of $2\theta$ that cause $\cos(2\theta)=-\dfrac{1}{2}$:

From this we see the solutions are $$\left\{ \begin{array}{ll} 2\theta &= \dfrac{2\pi}{3} + 2\pi k, \\ 2\theta &= \dfrac{4\pi}{3} + 2\pi k, \end{array} \right.$$ where $k$ is an integer. We want to isolate $\theta$ so now divide by $2$ in both equations to get $$\left\{ \begin{array}{ll} \theta &= \dfrac{\pi}{3} + \pi k, \\ \theta &= \dfrac{2\pi}{3} + \pi k. \end{array} \right.$$ To find the requested six solutions, we will pick particular values for $k$. First we pick $k=-1$ to get the solutions $$\left\{ \begin{array}{ll} \theta &= \dfrac{\pi}{3} - \pi = \dfrac{\pi}{3} - \dfrac{3\pi}{3} = -\dfrac{2\pi}{3}, \\ \theta &= \dfrac{2\pi}{3} - \pi = \dfrac{2\pi}{3} - \dfrac{3\pi}{3} = -\dfrac{\pi}{3}. \end{array} \right.$$ Notice neither of these solutions obey the inequality $0 \leq \theta \leq 2\pi$ so we do not consider them for the final answer. Now we plug in $k=0$ to get the solutions $$\left\{ \begin{array}{ll} \theta &= \dfrac{\pi}{3} + 0 =\dfrac{\pi}{3}, \\ \theta &= \dfrac{2\pi}{3} + 0 = \dfrac{2\pi}{3}. \end{array} \right.$$ Notice both of these solutions lie in the interval $0 \leq \theta \leq 2\pi$, so we will consider them for the final answer. Plug in $k=1$ to get the solutions $$\left\{ \begin{array}{ll} \theta &= \dfrac{\pi}{3} + \pi = \dfrac{\pi}{3} + \dfrac{3\pi}{3} = \dfrac{4\pi}{3}, \\ \theta &= \dfrac{2\pi}{3} + \pi = \dfrac{2\pi}{3} + \dfrac{3\pi}{3} = \dfrac{5\pi}{3}. \end{array} \right.$$ Notice both of these solutions lie in the interval $0 \leq \theta \leq 2\pi$. Finally plug in $k=2$ to get the solutions $$\left\{ \begin{array}{ll} \theta &= \dfrac{\pi}{3} + 2\pi = \dfrac{\pi}{3} = \dfrac{\pi}{3} + \dfrac{6\pi}{3} = \dfrac{7\pi}{3}, \\ \theta &= \dfrac{2\pi}{3} + 2\pi = \dfrac{2\pi}{3} + \dfrac{6\pi}{3} = \dfrac{8\pi}{3}. \end{array} \right.$$ Notice both of these solutions do not lie in the interval $0 \leq \theta \leq 2\pi$ (they are too large!). Therefore we have found four solutions in the interval $0\leq \theta \leq 2\pi$: $$\theta = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}.$$

Section 3.3 #28: Solve on the interval $0 \leq \theta \leq 2\pi$: $$5\csc(\theta)-3=2.$$
Solution: First isolate $\csc(\theta)$ by adding $3$ and dividing by $5$: $$\csc(\theta) = \dfrac{5}{5} = 1.$$ Now we observe the unit circle to find places where $\csc(\theta)=1$:

From this we see that the only solution is $\theta=\dfrac{\pi}{2}$.

Section 3.3 #36: Give a general formula for all solutions. List six solutions: $$\tan(\theta)=1.$$
Solution: Observe the unit circle for places where $\tan(\theta)=1$:

From this we see that the solutions are $$\left\{ \begin{array}{ll} \theta &= \dfrac{\pi}{4} + 2\pi k, \\ \theta &= \dfrac{5\pi}{4} + 2\pi k, \end{array} \right.$$ where $k$ is any integer. To find the requested six solutions, we will pick particular values for $k$. First we pick $k=-1$ to get the solutions $$\left\{ \begin{array}{ll} \theta &= \dfrac{\pi}{4} - 2\pi = \dfrac{\pi}{4} - \dfrac{8\pi}{4} = -\dfrac{7\pi}{4}, \\ \theta &= \dfrac{5\pi}{4} - 2\pi = \dfrac{5\pi}{4} - \dfrac{8\pi}{4} = -\dfrac{3\pi}{4}. \end{array} \right.$$ Now we plug in $k=0$ to get the solutions $$\left\{ \begin{array}{ll} \theta &= \dfrac{\pi}{4} + 0 =\dfrac{\pi}{4}, \\ \theta &= \dfrac{5\pi}{4} + 0 = \dfrac{5\pi}{4}. \end{array} \right.$$ Finally plug in $k=1$ to find the last two solutions: $$\left\{ \begin{array}{ll} \theta &= \dfrac{\pi}{4} + 2\pi = \dfrac{\pi}{4} + \dfrac{8\pi}{4} = \dfrac{9\pi}{4}, \\ \theta &= \dfrac{5\pi}{4} + 2\pi = \dfrac{5\pi}{4} + \dfrac{8\pi}{4} = \dfrac{13\pi}{4}. \end{array} \right.$$

Section 3.3 #44: Give a general formula for all solutions. List six solutions: $$\tan \left( \dfrac{\theta}{2} \right) = -1.$$
Solution: We must find values of $\dfrac{\theta}{2}$ for which $\tan \left( \dfrac{\theta}{2} \right)=-1$. Observe the unit circle:

From this we see the solutions are $$\left\{ \begin{array}{ll} \dfrac{\theta}{2} &= \dfrac{3\pi}{4} + 2\pi k, \\ \dfrac{\theta}{2} &= \dfrac{7\pi}{4} + 2\pi k, \end{array} \right.$$ where $k$ is an integer. First we isolate $\theta$ by multiplying each equation by $2$: $$\left\{ \begin{array}{ll} \theta &= \dfrac{3\pi}{2} + 4\pi k, \\ \theta &= \dfrac{7\pi}{2} + 4\pi k. \end{array} \right.$$ To find the requested six solutions, we begin by plugging in $k=-1$: $$\left\{ \begin{array}{ll} \theta &= \dfrac{3\pi}{2} - 4\pi = \dfrac{3\pi}{2} - \dfrac{8\pi}{2} = -\dfrac{5\pi}{2}, \\ \theta &= \dfrac{7\pi}{2} - 4\pi = \dfrac{7\pi}{2} - \dfrac{8\pi}{2} = -\dfrac{\pi}{2}. \end{array} \right.$$ Now we plug in $k=0$: $$\left\{ \begin{array}{ll} \theta &= \dfrac{3\pi}{2} + 0 = \dfrac{3\pi}{2}, \\ \theta &= \dfrac{7\pi}{2} + 0 = \dfrac{7 \pi}{2}. \end{array} \right.$$ Finally we will plug in $k=1$: $$\left\{ \begin{array}{ll} \theta &= \dfrac{3\pi}{2} + 4\pi = \dfrac{3\pi}{2} + \dfrac{8\pi}{2} = \dfrac{11\pi}{2}, \\ \theta &= \dfrac{7\pi}{2} + 4\pi = \dfrac{7\pi}{2} + \dfrac{8\pi}{2} = \dfrac{15\pi}{2}. \end{array} \right.$$