Question

if sinx+sin^2x=1 so find value of cos^12x+3cos^10x

Answer 1

i think here something is missing .

if your question
cos^12x+3cos^10x+3cos^8x+cos^6x
you can proceed this ,

sinx + sin^2x =1

sinx =1 - sin^2x =cos^2x

now,
cos^12x+3cos^10x +3cos^8x +cos^6x

=> (cos^4x)^3+3. (cos^4x)^2.cos^2x +3cos^4x.(cos^2x)^2+(cos^2x)^3

=>(sin^2x)^3+3sin^4x.cos^2x+3sin^2x.cos^4x +(cos^2x)^3

=>{sin^2x+ cos^2x}^3 = 1^3 =1

Answer 2

sin^2 x  + sin x - 1 = 0
sin x = [ √5 - 1] /2  = cos² x

cos⁴ x = (5+1-2√5)/4 = (3 - √5)/2
cos⁸ x = (9+5- 6√5)/4  = (7- 3√5)/2

3 cos¹° x + cos¹² x
  = cos⁸ x  * (3 cos² x + cos⁴ x)
  =  (7 - 3√5)/2 * [ 3 (√5 -1)/2 + (3 - √5)/2 ]
   = (7 - 3√5)/2 * √5
   = (7 √5 - 15)/2