solve: 2(sin^4x + cos^4x) = 1

Question

with the intervals : [-π ≤ x ≤ π ].

Answer 1

Given trigonometric equation is 2 ( sin ^ 4 ( x) + cos ^ 4 (x) ) = 1.

Intervals [ -∏ ≤ x ≤ ∏ ].

2 ( sin ^ 4 ( x) + cos ^ 4 (x) ) = 1.

 ( sin ^ 4 ( x) + cos ^ 4 (x) ) = 1 / 2

 [ ( sin ^ 2( x) ] ^ 2 + cos ^ 4 (x)  = 1 / 2

 [1 - cos^ 2( x) ] ^ 2 + cos ^ 4 (x)  = 1 / 2

 [1 + cos^ 4( x) - 2 cos^ 2( x) ] + cos ^ 4 (x)  = 1 / 2

2 cos^ 4( x) - 2 cos^ 2( x)  + 1 - 1 / 2 = 0

2 cos^ 4( x) - 2 cos^ 2( x)   + 1 / 2 = 0

4 cos^ 4( x) - 4 cos^ 2( x)  +  1 = 0

( 2cos^ 2( x) ) ^ 2  - ( 2 *2  cos ^ 2 (x) * 1) + 1 = 0

( 2cos^ 2( x) - 1 ) ^ 2 = 0

( 2cos^ 2( x) - 1 ) = 0

 2cos^ 2( x)  = 1

 cos^ 2( x)  = 1 / 2

 cos( x)  = 1 / √2  or  cos( x)  = - 1 / √2

Let us consider cos( x)  = 1 / √2 on [ - pie , pie ]

First in this case  x = ∏ / 4 or x =  -∏ / 4

Now we will consider cos( x)  = - 1 / √2  on [ - pie , pie ]

 in this case  x =- 3 ∏ / 4 or  x = 3∏ / 4

Hence the solutions will be

x = - ∏ / 4 , ∏ / 4 , -3 ∏ / 4 or 3∏ / 4.