Find solutions:

sin ( $\displaystyle \frac{1}{2}$ - 7x) $\displaystyle \times $ $\displaystyle \pi$ = cos(2-5x) $\displaystyle \times $ $\displaystyle \pi$ + sin8$\displaystyle \pi$

Interval - [$\displaystyle \frac{1}{2}$;1]

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What I did: 1) sin ( $\displaystyle \frac{1}{2}$ - 7x) = sin $\displaystyle \frac{1}{2}$cos7x + sin(7x)cos($\displaystyle \frac{1}{2}$)

2) cos(2-5x) = cos2cos5x + sin2sin5x

..and 3) sin8$\displaystyle \pi $ = sin2$\displaystyle \pi $

But what should I do now, should I multiply by $\displaystyle \pi $? Thank you.