Question

In the interval 0 degrees < x < 360 degrees, find the values of x for which cos x = 0.7252. Give your answers to the nearest degree.

Answer 1

Answer:

115° and 245°

Step-by-step explanation:

Given

cos x = - 0.4226

Since cos x < 0 then x is an angle in the second / third quadrants, thus

x =  (0.4226) = 65° ← related acute angle , thus

x = 180° - 65° = 115° ← angle in second quadrant

x = 180° + 65° = 245° ← angle in third quadrant

Answer 2

Answer:

$x=\{44\º, 317\º\}$

Step-by-step explanation:

The interval given is $(0, 360\º) \text{ or } (0, 2\pi)$

In exercises of this kind I usually use

$\cos \left(x\right)=a\quad \Rightarrow \quad \:x=\arccos \left(a\right)+360\º n, n \in \mathbb{Z}$

$\quad \:x=\arccos \left( 0.7252\right)+360\º n, n \in \mathbb{Z}$

And $\arccos \left( 0.7252\right) \approx 43.5^{\circ \:}$

But once we have the solution for cos in two different quadrants, I mean, Quadrant I and Quadrant IV angle.

To the nearest degree, we have

$x=\{44\º, 317\º\}$