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have no idea how to do 3b

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Hi u/FlounderTop9198,

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Find the roots of the polynomial in part b in terms of x (substitute u=x^2 and use the quadratic formula to find u, then replace u with x^2 to find x)

x is a root and cos(15) is a root, so at least one of the roots is cos(15) (something between -1 and 1. You can use the unit circle to narrow that range a little bit)
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Oh whoops. I gave you solution to c

For b, you know cos(4t)=8cos(t)^4 -8cos(t)^2 +1. You can write 16x^4 - 16x^2 +1 as 2(8x^4 -8x^2 +1) -1 (The -1 is because we needed to account for the extra one when we multiply by two)

From here it's similar to your earlier post.  If x is between -1 and 1, you can say x=cos(t) and use your trig identity to finish it off

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