The-most-general-solution-of-the-equation-sinx-cosx-min-a-R-1-a-2-4a-6-is-

Question Number 21050 by Tinkutara last updated on 10/Sep/17

The most general solution of the

equation \sin x + \cos x = \min_{a \in R} {1, a^{2} - 4 a + 6}

is

Answered by mrW1 last updated on 11/Sep/17

a^{2} - 4 a + 6 = a^{2} - 2 \times 2 \times a + 2^{2} + 2 = {(a - 2)}^{2} + 2 ⩾ 2 > 1

\Rightarrow \min_{a \in R} {1, a^{2} - 4 a + 6} = 1

\Rightarrow \sin x + \cos x = 1

\Rightarrow \sqrt{2} (\sin x \cos \frac{π}{4} + \sin \frac{π}{4} \cos x) = 1

\Rightarrow \sqrt{2} \sin (x + \frac{π}{4}) = 1

\Rightarrow \sin (x + \frac{π}{4}) = \frac{1}{\sqrt{2}}

\Rightarrow x + \frac{π}{4} = n π + {(- 1)}^{n} \frac{π}{4}

\Rightarrow x = n π + [{(- 1)}^{n} - 1] \frac{π}{4}

Commented by Tinkutara last updated on 12/Sep/17

Thank you very much Sir!