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What is the sum of all values of x (in degrees) satisfying sin^6 x−cos^6 x=1, where 0≤x≤360?
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\[\cos^{6} x = (1-\sin^{2}x)^{3} = 1-3\sin^{2} x +3\sin^{4} x -\sin^{6} x\] \[\rightarrow 2\sin^{6} x -3\sin^{4} x +3\sin^{2}x -2 = 0\] let u = sin(x) \[2u^{6}-3u^{4}+3u^{2} -2 = 0\] by graphing you can see roots of +-1 it factors: \[(u^{2}-1)(2u^{4}-u^{2}+2) = 0\] \[u^{2} = \sin^{2}x = 1\] \[\sin x = \pm 1\] \[x = 90, 270\] sum is 360
what about Sin^12x-Cos^12x =1
its done the same way...expand cos^12 in terms of sin then find roots of the 12th degree polynomial
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