Question

sin^8x+cos^8x = 97÷128​

Answer 1

Answer:

Step-by-step explanation:

let sin²x = k

sin^8x+cos^8x = sin^8x+ (1 - sin²x)^4 = k^4 +(1-K)^4

= 2k^4 - 4k³+6k²-4k +1 = 97÷128​

2k^4 - 4k³+6k²-4k +31/128 = 0

​k^4 - 2k³+3k²-2k +31/256 = 0