If sinx+sin²x=1 then the value of cos¹²x+3cos^10 x+3cos^8 x+cos^6 x+2 is:?

sinx + sin²x = 1 ⇒ sinx = cos²x , then
cos¹²x + 3cos¹ºx + 3cos^8 x + cos^6 x + 2
= (cos^6 x) (cos^6 x + 3cos⁴x + 3cos²x + 1) + 2
= (sin³x) (sin³x + 3sin²x + 3sinx + 1) + 2
= (sin³x) (sinx + 1)³ + 2
= (sinx + sin²x)³ + 2
= 1³ + 2
= 3

Hello,

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= = = = = = = = = = = = = = = = = = = = = =
► You state that:
   sin(𝑥) + sin²(𝑥) = 1
   sin(𝑥) = 1 – sin²(𝑥)

And we know that:
   1 = sin²(𝑥) + cos²(𝑥)
   1 – sin²(𝑥) = cos²(𝑥)

Thus:
   cos²(𝑥) = sin(𝑥)

► Now:
   cos¹²(𝑥) + 3·cos¹º(𝑥) + 3·cos⁸(𝑥) + cos⁶(𝑥) + 2
      = [cos⁴(𝑥)]³ + 3·[cos⁴(𝑥)]·cos²(𝑥) + 3·cos⁴(𝑥)·[cos²(𝑥)]² + [cos²(𝑥)]³ + 2
      = [cos⁴(𝑥) + cos²(𝑥)]³ + 2     ←←← Since 𝛼³+3𝛼²𝛽+3𝛼𝛽²+𝛽³=(𝛼+𝛽)³
      = [cos²(𝑥)·[cos²(𝑥) + 1] ]³ + 2    ←←← Factor by cos²(𝑥)
      = [sin(𝑥)·[sin(𝑥) + 1] ]³ + 2    ←←← Since cos²(𝑥)=sin(𝑥)
      = [sin²(𝑥) + sin(𝑥)]³ + 2     ←←← Expand
      = 1³ + 2      ←←← Since sin(𝑥)+sin²(𝑥)=1
      = 3           ◄◄◄ANSWER

Regards,
Dragon.Jade :-)