Question

Aratati ca sin x+ rad(3)*cos x<=2 , oricare x apartine R

Answer 1

$sinx+\sqrt3cosx \leq 2$

$f(x) = sinx+\sqrt3cosx$

$f'(x) = 0 => -\sqrt3sinx+cosx=0$

$-\sqrt3sinx=-cosx | (-1) => \sqrt3sinx=cosx | :cosx$

$\sqrt3tgx=1 => tgx = \frac{1}{\sqrt3}$

$x=\frac{\pi}{6}; \frac{7\pi}{6}$

$f''(x) = -\sqrt3cosx-sinx$

$f''(\frac{\pi}{6}) = -\sqrt3\frac{\sqrt3}{2} -\frac{1}{2} < 0 =>$ f are maxim la $x = \frac{\pi}{6}$

Valoarea maxima pentru $f(\frac{\pi}{6}) = \sqrt3* \frac{\sqrt3}{2} + \frac{1}{2} = 2$

Answer 2

1*sinx+√3cosx≤2 | impartim toata relatia cu 2

(1/2) *sinx+(√3/2) * cosx≤1

cos(π/3)sinx+sin(π/3)cosx≤1

sin(x+π/3)≤1

fie α=x+π/3

-1≤sinα≤1, adevarat ∀α∈R⇒adevarat ∀x=α-π/3