Question

Just the answers please.
1: http://imgur.com/blG548a
2: http://imgur.com/xnNR9CX
3: http://imgur.com/UYVfRkd

Answer 1

Answer on Question #59342 – Math – Trigonometry

Question

1. Which value is a solution for the equation $\tan \frac{x}{2} = 0$?

$$\begin{array}{l} \pi \\ 3\pi \\ \hline 2 \\ \pi \\ \hline 2 \\ 2\pi \\ \end{array}$$

Solution

$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$, then $\tan \frac{x}{2} = 0$ when the numerator is equal to zero. We know $\sin(\alpha) = 0$ when $\alpha = \pi \cdot n, n = 0, \pm 1, \pm 2, \ldots$. Then we equate $\frac{x}{2}$ to $\pi \cdot n$ and solve equation:

$$\frac{x}{2} = \pi \cdot n$$ $$x = 2\pi \cdot n,$$

hence $2\pi$ is a solution to the equation.

Answer: $2\pi$

Question

2. The value $\frac{5\pi}{4}$ is a solution for the equation $3\sqrt{2} \sec \theta + 7 = 1$?

True

False

Solution

Let's check if $\frac{5\pi}{4}$ is a solution. Substitute it into the equation:

$$3\sqrt{2} \sec \theta + 7 = 1$$ $$3\sqrt{2} \sec \left(\frac{5\pi}{4}\right) + 7 = 1$$

We know that $\sec \alpha = \frac{1}{\cos \alpha}$.

$$3\sqrt{2} \frac{1}{\cos \left(\frac{5\pi}{4}\right)} + 7 = 1,$$ $$\cos \left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2},$$ $$3\sqrt{2} \frac{1}{-\frac{\sqrt{2}}{2}} + 7 = 1,$$ $$3 \sqrt {2} \cdot \left(- \frac {2}{\sqrt {2}}\right) + 7 = 1,$$ $$- 6 + 7 = 1,$$

which is true.

Answer: True.

Question

3. There is no solution to the equation $cscx = -1$.

False

True

Solution

Let's solve equation $cscx = -1$. We know $csc \propto = \frac{1}{\sin \alpha}$, then we can rewrite the equation as

$$\frac {1}{\sin x} = - 1,$$ $$\sin x = - 1,$$ $$x = \frac {3 \pi}{2} + 2 \pi n, n = 0, \pm 1, \pm 2, \dots$$

We can see a solution exists.

Answer: False.

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