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

Answer:

The correct option is;

a- 45° only

Step-by-step explanation:

The given equation can be presented as follows;

[tex]\dfrac{cos (2 \cdot x)}{cosine (x) + sin (x)} = 0[/tex]

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;

[tex]\dfrac{cos (2 \cdot x)}{cosine (x) + sin (x)} = \dfrac{(cos(x) - sin(x)) \times (cos(x) + sin(x))}{(cos (x) + sin (x))} =0[/tex]

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

[tex]\dfrac{(cos(x) - sin(x)) \times (cos(x) + sin(x))}{(cos (x) + sin (x))} = cos(x) - sin(x) = 0[/tex]

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°

Answer:

A. 45 only

Step-by-step explanation:

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