The sum of the roots of the equation $4\cos^3 (\pi + x) - 4 \cos^2 (\pi - x) + \cos (\pi + x) - 1= 0$ in the interval $[0, 320]$ is p$\pi$ where p is equal to

Applying reduction formulas, cosine of pi plus x is negative cosine of x and cosine of pi minus x is negative cosine of x. Substituting these into the equation yields negative 4 cosine cubed of x minus 4 cosine squared of x minus cosine of x minus 1 equals 0, which rearranges to 4 cosine cubed of x plus 4 cosine squared of x plus cosine of x plus 1 equals 0. Factoring by grouping gives (cosine of x plus 1)(4 cosine squared of x plus 1) equals 0; since 4 cosine squared of x plus 1 is always positive, cosine of x must equal negative 1, meaning x equals (2k minus 1) times pi for integer values of k. The roots in the interval from 0 to 320 are pi, 3pi, 5pi, and so on up to 101pi; applying the sum of an arithmetic progression, this sum evaluates to 2601pi, so p equals 2601.

• Thanks
• Wrong explanation