Question

hey buddies please solve this​

Answer 1

Given that

sinx + sin²x = 1

so,

sin²x + sinx - 1 = 0

taking y = sinx

therefore,

y² + y -1 = 0

y = [-1 ± √(1+4)]\ 2

y = (√5) - 1 \ 2

so, sinx = √5 - 1 \2

sin²x = 3 - √5

then ,

cos²x = 1 - sin²x

cos²x = 1 - 3 + √5

cos²x = √5 - 2.

Now,

cos¹²x + 3cos^10x + 3 cos^8x + cos^6x + 2cos⁴x + 2 cos²x - 1 = ?

(cos²x)^6 + 3(cos²x)^5 + 3(cos²x)⁴ + (cos²x)³ + 2(cos²x)² + 2cos²x - 1 = ?

Now put cos²x = √5 - 2,

Solve further... You will get the answer...

I hope this may help you...

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