CPM Homework Help : A2C Problem 9-155

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Home > A2C > Chapter 9 > Lesson 9.3.2 > Problem 9-155

9-155.

For each equation, find two solutions $0 \lt x \lt 2$ , which make the equation true. No calculator necessary.

$\operatorname { cos } x = - \frac { 1 } { 2 }$
Hint (a):
$\text{cos}\left(\frac{\pi}{6}\right)=\frac{1}{2}$
Step 1(a):
What angle has the cosine of negative one half? Think of 30°, 60° right triangles.
x = 120° or 240°
Step 2(a):
$120\left(\frac{\pi}{180}\right)\,\,\,\,\,\,\,240\left(\frac{\pi}{180}\right)$
Now convert the angles into radians.
Answer (a):
$\textit{x}=\frac{2\pi}{3},\,\frac{4\pi}{3}$

$\operatorname{tan} x =\frac { \sqrt { 3 } } { 3 }$
Hint (b):
See part (a).

$\operatorname{sin} x = 0$
Hint (c):
If sin(x) = 0, y = 0
Answer (c):
$x = 0$

$\operatorname{cos} x =\frac { \sqrt { 2 } } { 2 }$
Hint (d):
Remember the side ratios of $45°$ right triangles.

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