if the maximum value of 4sin^2 x+3cos^2 x+sin (x/2)+cos (x/2)+3 is a+(√b) then find a+b


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Question Number 146876 by gsk2684 last updated on 16/Jul/21

$i f t h e m a x i m u m v a l u e o f$ $4 \sin^{2} x + 3 \cos^{2} x + \sin \frac{x}{2} + \cos \frac{x}{2} + 3$ $i s a + \sqrt{b} t h e n f i n d a + b$
	Answered by liberty last updated on 16/Jul/21
	$f(x)=4sin^2 x+3−3sin^2 x+3+sin ((x/2))+cos ((x/2)) f(x)=sin^2 x+sin ((x/2))+cos ((x/2))+6 f(x)=sin^2 x+sin ((x/2))+cos ((x/2))+6 f ′(x)=sin 2x+(1/2)cos ((x/2))−(1/2)sin ((x/2))=0 sin 2x = (1/2)sin ((x/2))−(1/2)cos ((x/2)) 2sin 2x −sin ((x/2))+cos ((x/2))=0 let (x/2)=u →x=2u 2sin 4u+(cos u−sin u)=0 4sin 2u (cos^2 u−sin^2 u)+(cos u−sin u)=0 (cos u−sin u){4sin 2u(cos u+sin u)+1}=0 (•) cos u−sin u=0 cos u=sin u ⇒tan ((x/2))=1 x=(π/2) f((π/2))= 1+6 +((√2)/2)+((√2)/2)=7+(√2) f_(max) = 7+(√2) ≈ 8.414213 then { ((a=7)),((b=2)) :} ⇒a+b=9$
	$f (x) = 4 \sin^{2} x + 3 - 3 \sin^{2} x + 3 + \sin (\frac{x}{2}) + \cos (\frac{x}{2})$ $f (x) = \sin^{2} x + \sin (\frac{x}{2}) + \cos (\frac{x}{2}) + 6$ $f (x) = \sin^{2} x + \sin (\frac{x}{2}) + \cos (\frac{x}{2}) + 6$ $f^{'} (x) = \sin 2 x + \frac{1}{2} \cos (\frac{x}{2}) - \frac{1}{2} \sin (\frac{x}{2}) = 0$ $\sin 2 x = \frac{1}{2} \sin (\frac{x}{2}) - \frac{1}{2} \cos (\frac{x}{2})$ $2 \sin 2 x - \sin (\frac{x}{2}) + \cos (\frac{x}{2}) = 0$ $l e t \frac{x}{2} = u \to x = 2 u$ $2 \sin 4 u + (\cos u - \sin u) = 0$ $4 \sin 2 u (\cos^{2} u - \sin^{2} u) + (\cos u - \sin u) = 0$ $(\cos u - \sin u) {4 \sin 2 u (\cos u + \sin u) + 1} = 0$ $(∙) \cos u - \sin u = 0$ $\cos u = \sin u \Rightarrow \tan (\frac{x}{2}) = 1$ $x = \frac{π}{2}$ $f (\frac{π}{2}) = 1 + 6 + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 7 + \sqrt{2}$ $f_{m a x} = 7 + \sqrt{2} \approx 8.414213$ $t h e n {\begin{cases} a = 7 \\ b = 2 \end{cases} \Rightarrow a + b = 9$
	Commented by gsk2684 last updated on 16/Jul/21

	$t h a n k y o u$
	Commented by liberty last updated on 16/Jul/21

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