Differences of powers

This is possible only when abs(sin x) = 1 => x = 90 or 270.

sin^12(x)−cos^12(x)=1 or, sin^12(x) = 1+cos^12(x) or, (sin^6(x))^2 = 1+(cos^6(x))^2 or, 1 - (cos^6(x))^2 = 1+(cos^6(x))^2 or, 2(cos^6(x))^2 = 0 or, (cos^6(x))^2 = 0 or, cos^6(x) = 0 or, cos(x) = 0 or, x = 90 and 270 or, Summation of all the values of x = 90+270 = 360 (in degrees.)

$\sin^{12} x$ and $\cos^{12} x$ both has a range of [0,1]

Therefore, in order for the equation to be satisfied, $\sin^{12} x$ must be equal to 1 and $\cos^{12} x$ must be equal to 0. This can only happen at $x=90,270$

Note that $sin^{12}x$ and $cos^{12}x$ are nonnegative and are in the range $[0,1]$ . Since $sin^{12}x-cos^{12}x=1$ , $sin^{12}x=1$ . Hence $sin x=\pm 1$ , so $x=90^\circ$ or $270^\circ$ . The sum of all possible values of $x$ is $90+270=\boxed{360}$ .

Since the maximum value of $sin(x)$ and $cos(x)$ is $1$ and since the twelfth power of either $cos(x)$ or $sin(x)$ is strictly positive, $\Rightarrow sin^{12}(x) = 1$

$\Rightarrow x = 90^{\circ}, 270^{\circ}$ . Thus sum of all values of x is $360^{\circ}$

We add 2cos^12 x to both sides so that we have sin^12 x +cos^12 x=1+2cos^12 x . The RHS is > or = to 1, and the maximum of the LHS is 1 (because sin^2 x +cos^2 x =1, and both sin^12 and cos^12 are less than sin^2 and cos^2). Therefore, we solve the equation sin^12 x +cos^12 x=1, and this only has solutions when either sin^12 x or cos^12 x is equal to 1, because sin^12 x or cos^12 x<sin^2 x+cos^2 x for all other cases. sin^12 x=1 only at 90 and 270 degrees, and cos^12 x=1 at 0, 180, and 360 degrees. However, the cosine gives extraneous solutions, so we only take the ones from the sine, which are 90+270=360