The general term in the expansion of
(‌1∕x+xsin‌x)10 is given by:
‌Tr+1=‌10Cr(‌1∕x)10−r(xsin‌x)r=‌10Crxr−(10−r)sin‌rx
‌Tr+1=‌10Crx2r−10sin‌rx
We are asked to find the coefficient of the 6th term, which means we need to find the value of r when
r+1=6, or
r=5.
Substituting
r=5 into the general term formula, we get:
T6=‌10C5x0sin‌5x=‌10C5sin‌5x We are told that this coefficient is equal to
7‌, so we have the equation:
‌10C5sin‌5x=7‌=‌Now, we need to find the value of
x that satisfies this equation. We can use a calculator to find the value of
‌10C5=252.
Therefore:
‌252sin‌5x=‌‌sin‌5x=‌=‌‌sin‌x=‌5√‌=‌The principal value of
x for which
sin‌x=‌ is
x=30∘.