Find-the-maximum-value-of-3sin-2-x-8cosx-5-

Question Number 213643 by Abdullahrussell last updated on 12/Nov/24

F i n d t h e m a x i m u m v a l u e o f

3 {s i n}^{2} x - 8 c o s x + 5 = ?

Answered by Berbere last updated on 12/Nov/24

{s i n}^{2} (x) = 1 - {c o s}^{2} (x)

t = c o s (x); t \in [- 1, 1]

s t u d y t \to f (t) = 3 (1 - t^{2}) - 8 t + 5 t

Answered by golsendro last updated on 12/Nov/24

F (x) = 3 (1 - \cos^{2} x) - 8 \cos x + 5

F (x) = - 3 \cos^{2} x - 8 \cos x + 8

\cos x = 1 \Rightarrow F_{1} = - 3 - 8 + 8 = - 3

\cos x = - 1 \Rightarrow F_{2} = - 3 + 8 + 8 = 13

\cos x = - \frac{(- 8)}{2 . (- 3)} = - \frac{4}{3} (rejected)

∴ \max value = 13

Answered by a.lgnaoui last updated on 12/Nov/24

let f (x) = 3 \sin^{2} x - 8 \cos x + 5

{\begin{cases} Max ({3 \sin}^{2} x, - 8 \cos x) = + 8 [x = (2 k + 1) π] \\ \Rightarrow f (x) ⩽ 13 \end{cases}

So \boldsymbol Max (\boldsymbol f (\boldsymbol x)) = 13

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