Find the maximum value of the function $sinx(1+cosx)$ is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{3\sqrt{3}}{4}$
$y=\sin x(1+\cos x)$
$y'=\cos x(1+\cos x)-\sin x\sin x$
$=\cos x+\cos^2x-\sin^2x=0$
$2\cos^2x+\cos x-1=0$
$(2\cos x-1)(\cos x+1)=0$
$\cos x=\frac{1}{2},\cos x=-1$
so $y_{max}=\frac{\sqrt{3}}{2}(1+\frac{1}{2})=\frac{3\sqrt{3}}{4}$