Question

Solve `cos^2x+cosx-1=0` ?

Answer 1

Given `cosx+ cos^2x=1 .......(1)`

`=>cosx=1-cos^2x`

`=>cosx=sin^2x`

Again from (1)

`cos^2x+cosx-1=0`

And `cosx=1/2(-1+sqrt(1^2-4*1*(-1)))`
`=>cosx=1/2(-1+sqrt5)`

other root

`cosx=1/2(-1-sqrt5)<-1->"not possible"`

So

`sin^12x +sin^10x + 3sin^8x +sin^6x`

`=(sin^2x)^6 +sin^10x + 3(sin^2x)^4 +sin^6x`

`= (cosx)^6 +sin^10x + 3(cosx)^4+sin^6x`

`= cos^6x +sin^6x + 3cos^4x +sin^10x`

`= (cos^2x +sin^2x)^3-3cos^2xsin^2x(cos^2x+sin^2x) + 3cos^4x +(sin^2x)^5`

`=1^3-3cos^2xcosx + 3cos^4x +cos^5x`

`=1-3cos^3x(1- cosx) +cos^5x`

`=1-3cos^3x*cos^2x +cos^5x`

`=1-2cos^5x=1-2(cos^2x)cosx`

`=1-2(1-cosx)^2*cosx`

`=1-2cosx+4cos^2x-2cos^3x`

`=1-2cosx+4cos^2x-2cos^2x*cosx`

`=1-2cosx+4cos^2x-2(1-cosx)*cosx`

`=1-2cosx+4cos^2x-2cosx+2cos^2x`

`=1-4cosx+6cos^2x`

`=1-4cosx+6(1-cosx)`

`=7-10cosx`

`=7-10xx1/2(-1+sqrt5)`

`=12-5sqrt5`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

When the given expression
`= sin^12x +3sin^10x + 3sin^8x +sin^6x`

`=(sin^2x)^6 +3sin^10x + 3(sin^2x)^4 +sin^6x`

`= (cosx)^6 +3sin^10x + 3(cosx)^4+sin^6x`

`= cos^6x +sin^6x + 3cos^4x +3sin^10x`

`= (cos^2x +sin^2x)^3-3cos^2xsin^2x(cos^2x+sin^2x) + 3cos^4x +3(sin^2x)^5`

`=1^3-3cos^2xcosx + 3cos^4x +3cos^5x`

`=1-3cos^3x(1- cosx) +3cos^5x`

`=1-3cos^3x*cos^2x +3cos^5x`

`=1-3cos^5x +3cos^5x`

`=1`

Answer 2

`12 - 5 sqrt[5]`

Explanation:

Solving

`cos^2x+cosx-1=0` we obtain

`cosx = 1/2 (-1 pm sqrt[5])`.

and

`sinx = sqrt(1/2(-1+sqrt(5)))`

Putting this results into the big equation

`sin^12x+cdots+sin^6x` we obtain the answer.

Example

`(sqrt[1/2 (-1 + sqrt[5])])^16 = 1/2(47-21sqrt(5))`

so the answer is

`12 - 5 sqrt[5]`

If the big equation were instead

`sin^12x + 3 sin^10x + 3 sin^8x + sin^6x`

the result would be simply `1`