An angle measuring (468n)° is in standard position. for which value of n will the terminal side fall on the x-axis?

Given: Measure of angle =  (468 n)°

To Find: The value of n for  which  the terminal side fall on the x-axis

Solution:

Angle =(468 n)°

for , n=1

Angle = 468°= 360°+108°→108°> 90°(Second quadrant)

for, n=2

Angle = 468°×2=936°=720°+216°→216°> 180°(Third quadrant)

For, n=3

Angle = 468°× 3 = 1404°=1080°+324°→324°>270°(Fourth quadrant)

For, n=4

Angle = 468°× 4 = 1872°=1800°+72°→72°(First quadrant)

For, n=5

Angle = 468°× 5 = 2340°=180°×13→Falls on X axis.

Alternative Method:

(468 n)°=90° k

n=, so if k is a multiple of 26 , we get different values of n.

Smallest value of n for which  terminal side fall on the x-axis is when (k=26, n=5), (k=  52, n=10),......

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.