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Question Number 18892 by Tinkutara last updated on 01/Aug/17

If the equation \sin 6 x + \cos 4 x = - 2 have

a family of nonnegative solutions x_{k}^{'} s,

where 0 ⩽ x_{1} < x_{2} < x_{3} < \dots . < x_{k} < x_{k + 1}

\dots . ., then the value of \frac{1}{π} \underset{k = 1}{\sum^{1000}} ∣ x_{k + 1} - x_{k} ∣ is

Answered by ajfour last updated on 01/Aug/17

\sin 6 x + \cos 4 x = - 2

\Rightarrow \sin 6 x = - 1 and \cos 4 x = - 1

\Rightarrow 6 x = 2 n π + \frac{3 π}{2}; n ⩾ 0 (here)

or x = \frac{n π}{3} + \frac{π}{4} \dots (i)

And 4 x = 2 m π + π; m ⩾ 0

or x = \frac{m π}{2} + \frac{π}{4} \dots (ii)

since both (i) and (ii) together

must be true,

x - \frac{π}{4} = \frac{n π}{3} = \frac{m π}{2}

or x - \frac{π}{4} = k π

x_{k + 1} - x_{k} = π

\Rightarrow \frac{1}{π} \underset{k = 1}{\sum^{1000}} ∣ x_{k + 1} - x_{k} ∣= \frac{1000 π}{π} = 1000 .

Commented by ajfour last updated on 01/Aug/17

is it correct ?

Commented by Tinkutara last updated on 01/Aug/17

Thank you very much ajfour Sir!