Question

Just the answers please.
1: http://imgur.com/19HaJwc
2: http://imgur.com/tNifnIe
3: http://imgur.com/kLcfAIu

Answer 1

Answer on Question #59341 – Math – Trigonometry

Question

Solve $\sin(x) (\sin x + 1) = 0$,

$$x = \pm \pi n, x = \frac{\pi}{2} \pm 2\pi n$$ $$x = \pm \pi n$$ $$x = \pm \pi n, x = \frac{3\pi}{2} \pm 2\pi n$$ $$x = \frac{\pi}{2} \pm 2\pi n, x = \frac{3\pi}{2} \pm 2\pi n$$

Solution

$$\sin(x) (\sin x + 1) = 0$$ $$\begin{cases} \sin x = 0, \\ \sin x + 1 = 0, \end{cases}$$ $$\begin{cases} \sin x = 0, \\ \sin x = -1, \end{cases}$$ $$\begin{cases} x = \pm \pi n, \\ x = \frac{3\pi}{2} \pm 2\pi n, \end{cases}$$

$n$ is integer.

Answer: $x = \pm \pi n, x = \frac{3\pi}{2} \pm 2\pi n$.

Question

Solve on the interval $[0, 2\pi)$:

$$1 + \cos \theta = \frac{\sqrt{3} + 2}{2}$$ $$\frac{7\pi}{6}, \frac{11\pi}{6}$$ $$\frac{\pi}{6}, \frac{11\pi}{6}$$ $$\frac{\pi}{6}, \frac{5\pi}{6}$$ $$\frac{\pi}{3}, \frac{5\pi}{3}$$

Solution

$$1 + \cos(\theta) = \frac{\sqrt{3} + 2}{2}, \quad 0 < \theta < 2\pi,$$ $$1 + \cos(\theta) = \frac{\sqrt{3}}{2} + 1, \quad 0 < \theta < 2\pi,$$ $$\cos(\theta) = \frac{\sqrt{3}}{2}, \quad 0 < \theta < 2\pi.$$ $$\theta = \frac{\pi}{6} \text{ or } \theta = \frac{11\pi}{6}.$$

**Answer:** $\frac{\pi}{6}, \frac{11\pi}{6}$.

Question

Solve on the interval $[0, 2\pi)$

$$2\sec(x) + 4 = 0$$ $$\frac {\pi}{3}, \frac {5 \pi}{3}$$ $$\frac {7 \pi}{6}, \frac {11 \pi}{6}$$ $$\frac {2 \pi}{3}, \frac {4 \pi}{3}$$ $$\frac {\pi}{6}, \frac {5 \pi}{6}$$

Solution

$$\begin{array}{l} 2 \sec (x) + 4 = 0, 0 < x < 2\pi \\ \sec (x) + 2 = 0, 0 < x < 2\pi \\ \frac {1}{\cos (x)} + 2 = 0, 0 < x < 2\pi \\ \cos (x) = - \frac {1}{2}, 0 < x < 2\pi \\ x = \frac {2 \pi}{3} \text{ or } x = \frac {4 \pi}{3}. \end{array}$$

Answer: $\frac{2\pi}{3}, \frac{4\pi}{3}$.

www.AssignmentExpert.com