Help me to solve this trig

[ 2(cos^6x + sin^6x) - sinxcosx ] / ( √2 - 2sinx ) = 0

[ 2(cos^6x + sin^6x) - sinxcosx ] / ( √2 - 2sinx ) = 0
We know :
√2 - 2sinx # 0
<=> sinx # √2/2

<=> x # π/4 + k2π ( k of Z )
<=> x # π - π/4 + k2π => x # 3π/4 + k2π

=> 2(cos^6x + sin^6x) - sinxcosx = 0
<=> 2(cos^6x + 3sin²xcos^4x + 3sin^4xcos²x + sin^6x - 3sin²xcos^4x - 3sin^4xcos²x ) - sinxcosx = 0
<=> 2[(sin²x + cos²x)^3 - 3sin²xcos^4x - 3sin^4xcos²x ] - sinxcosx = 0
<=> 2[1 - 3sin²xcos²x(sin²x + cos²x) ] - sinxcosx = 0
<=> 2[1 - 3sin²xcos²x ] - sinxcosx = 0
<=> 2 - 6sin²xcos²x - sinxcosx = 0
<=> -6sin²xcos²x - sinxcosx + 2 = 0
Change sinxcosx = t
=> -6t² - t + 2 = 0

<=> t =1/2
<=> t = -2/3

<=> sinxcosx = 1/2
<=> sinxcosx = -2/3

<=> 2sinxcosx = 1
<=> 2sinxcosx = -4/3

<=> sin2x = 1 ( choose )
<=> sin2x = -4/3 ( don't choose because sin2x not in [-1 ; 1 ] )

<=> 2x = π/2 + k2π ( k of Z )
<=> x = π/4 + kπ

Since the expression is equal to zero, then you need to set the numerator equal to zero.

a/b = 0
means that a=0 AND b is not equal to zero.

If you solve

2(cos^6x + sin^6x) - sinxcosx = 0

Then you will solve the whole thing
(you will need to check that a solution does not cause the denominator to also equal zero, in which case the expression could be undefined.)

2(cos^6x + sin^6x) = sinxcosx

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If you are in the complex field, you can rewrite the left side as

2(cos^6x - i^6sin^6x)
a difference of powers

a^6 + b^6 =
a^6 - i^6b^6
=(a - ib)(a^5 +a^4ib +a^3i^2b^2 + a^2i^3b^3 + ai^4b^4 + i^5b^5)
=(a - ib)(a^5 + i(a^4b) - a^3b^2 -i(a^2b^3) + ab^4 + ib^5)
=(a - ib)( (a^5 - a^3b^2 + ab^4) + i(a^4b + b^5) )

Which looks even worse once you slip back the cos for a and sin for b.
However, it is just the product of two complex numbers, which can be solved in polar coordinates (not easily, but...)

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