Question

Find the maximum value of 4sin^2x + 3cos^x + sinx/2 + cosx/2 ? ​

Answer 1

Answer:

Maximum value = 4 + √2.

✔️Hope it will help you.✔️

Answer 2

Answer:

Maximum value of $4 \sin ^{2 x}+3 \cos ^{x}+\sin \frac{x}{2}+\cos \frac{x}{2}=4+√2$

Given Data:

$4 \sin ^{2 x}+3 \cos ^{x}+\sin \frac{x}{2}+\cos \frac{x}{2}$

To Find:

Maximum value?

Step 1:

$I=4 \sin 2 x+3(1-\sin 2 x)+\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)$

$I=4 \sin 2 x+3(1-\sin 2 x)+\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)$

$I =4 \sin 2 x+3-3 \sin 2 x+\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)$

Step 2:

$I=4 \sin 2 x+3-3 \sin 2 x+\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)$

$I=\sin 2 x+3+\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)$

$I=\sin 2 x+3+\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)$

$\left.I=\sin 2 x+3+2-\sqrt{[12-\sqrt{\sin \left(x^{2}\right)}}+\cos \left(x^{2}\right) \times 12\right$

Step 3:

$I=\sin 2 x+3+2\left[12 \sin \left(x^{2}\right)+\cos \left(x^{2}\right) 12\right]$

$=3+\sin 2 x+2-\sin \left[\pi 4+x^{2}\right]$

$=3+\sin 2 x+2 \sin \left[\pi^{4}+x^{2}\right] \sin \left(\pi^{4}+x^{2}\right)$

$=3+\sin 2 x+2 \sin \left[\pi^{4}+x^{2}\right] \sin \left(\pi^{4}+x^{2}\right)$

$=\sin \left(\pi^{4}+x^{2}\right)=1$

Step 4:

When$x=\pi 2 x=\pi^{2}$

i.e. $\sin \left(\pi^{4}+\pi^{4}\right)=\sin \left(\pi^{2}\right)=1 \sin \left(\pi^{4}+\pi^{4}\right)=\sin \left(\pi^{2}\right)=1$

Step 5:

Maxi $=x=\pi 2 I=x=\pi^{2}$

I (at $x=\pi^{2}$)  $=3+\sin 2\left(\pi^{2}\right)+\sqrt{2}-I\left(a t x=\pi^{2}\right)=3+\sin 2\left(\pi^{2}\right)+\sqrt{2}$

$=3+1+\sqrt{2}$

$=4+\sqrt{2}$