How do you find the value of $f(x)$ for this trig function satisfying all values of $x$? If $ f(x) = 3[\sin^4(\frac{3\pi}{2} - x) + \sin^4(3\pi+x)] -2[\sin^6(\frac{\pi}{2} + x) + \sin^6(5\pi-x)] $ then, for all permissible values of $x$, $f(x)$ is:-
Here's how I attempted it-
$ f(x) = 3[\sin^4(\frac{3\pi}{2} - x) + \sin^4(3\pi+x)] -2[\sin^6(\frac{\pi}{2} + x) + \sin^6(5\pi-x)] $
$ f(x) = 3[\cos^4x + \sin^4x] -2[\cos^6x + \sin^6x] $
$ f(x) = 3\cos^4x-2\cos^6x+ 3\sin^4x - 2\sin^6x $
$ f(x) = \cos^4x(3-2\cos^2x)+ \sin^4x(3 - 2\sin^2x) $
$ f(x) = \cos^4x(1+2\sin^2x)+ \sin^4x(1+2\cos^2x) $
$ f(x) = \cos^4x+2\sin^2x\cos^4x+ \sin^4x+2\cos^2x\sin^4x $
$ f(x) = (\sin^2x)^2+ (\cos^2x)^2 +2\sin^2x\cos^2x(\cos^2x+\sin^2x) $
$ f(x) = (\sin^2x+\cos^2x)^2$
I don't know what to do next. Thanks in advance to anyone who's willing to help. The final answer should come up to '$-1$'.(According to the textbook).

Since $$\sin \left(\frac{3\pi}{2} - x\right) = \cos x \implies \sin^4 \left(\frac{3\pi}{2} - x\right) = \cos^4 x$$ and $$\sin^4 (3\pi + x) = \sin^4 x$$
We have $f(x) = f(x) = 3(\sin ^4 x + \cos ^4 x)-2[\sin^6(\frac{\pi}{2} + x) + \sin^6(5\pi-x)]$. Let's work on simplifying the second part of the function now. We have $$\sin \left(\frac{\pi}{2} + x\right) = \cos x \implies \sin^6 \left(\frac{\pi}{2} + x\right) = \cos^6 x$$ and $$\sin^6 (5\pi-x) = \sin^6 x$$ so that your function becomes $$f(x) = 3(\sin^4 x + \cos^4 x) - 2(\sin^6 x + \cos^6 x)$$
But we know that $$\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x$$ and $$\sin^4 x + \cos^4 x = 1 - \frac{1}{2}\sin^2 2x = 1 - 2\sin^2 x \cos^2 x$$ Putting this all together yields $$f(x) = 3 - 6 \sin^2 x \cos^2 x - 2 + 6\cos^2 x \sin^2 x = 1$$
