Let StartFraction cosine (2 x) Over cosine (x) + sine (x) EndFraction = 0 where 0° ≤ x ≤ 180°. What are the possible values for x? a- 45 only
b- 135 only
c- 45 or 225
d- 135 or 315

*NEED HELP ASAP

The correct option is;

a- 45° only

Step-by-step explanation:

The given equation can be presented as follows;

We have that cos(2·x) = cos²(x) - sin²(x)

Also, we have, by the difference of two squares, the following relation;

cos²(x) - sin²(x) = (cos(x) - sin(x))(cos(x) + sin(x))

Therefore, the given equation can be written as follows;

Crossing the common term, (cos(x) + sin(x)), in the numerator and the denominator, we have;

From cos(x) - sin(x) = 0, we have;

Adding sin(x) to both sides of the equation

cos(x) - sin(x) + sin(x) = 0 + sin(x)

cos(x) = sin(x)

Therefore, the opposite leg and the adjacent leg of the right triangle formed with reference to the angle x are equal

∴ x = 90/2 = 45° only for 0° ≤ x ≤ 180°

d. it is a convex pentagon because it has five sides and none of the sides would extend into the inside of the polygon.

step-by-step explanation:

the answer is two. hope this !