Find All Solutions Of$\cos ^ { 2 } \pi x - \sin ^ { 2 } \pi x = 0$

Question

Examlex · Accepted Answer

The Answer is

\cos ^ { 2 } \pi x - \sin ^ { 2 } \pi x = 0

\Leftrightarrow

2 \cos ^ { 2 } \pi x - 1 = 0

\Leftrightarrow

\cos ^ { 2 } \pi x = \frac { 1 } { 2 }

\Leftrightarrow

\cos \pi x = \pm \frac { \sqrt { 2 } } { 2 }

, which has general solutions

\pi x = \frac { \pi } { 4 } + 2 k \pi

,

\pi x = \frac { 7 \pi } { 4 } + 2 k \pi

,

\pi x = \frac { 3 \pi } { 4 } + 2 k \pi

, and

\pi x = \frac { 5 \pi } { 4 } + 2 k \pi

, these can be written

x = \frac { 1 } { 4 } + 2 k

,

x = \frac { 7 } { 4 } + 2 k

,

x = \frac { 3 } { 4 } + 2 k

,

x = \frac { 5 } { 4 } + 2 k

, or briefly as

x = \frac { 1 } { 4 } + \frac { k } { 2 }

, where

k

is an integer.

Find All Solutions Of $\cos ^ { 2 } \pi x - \sin ^ { 2 } \pi x = 0$