The-number-of-solutions-of-the-equation-sin-3-x-3sinxcos-2-x-2cos-3-x-0-in-pi-4-pi-4-is-

Question Number 18524 by Tinkutara last updated on 23/Jul/17

The number of solutions of the equation

\sin^{3} x - 3 \sin x \cos^{2} x + {2 \cos}^{3} x = 0 in

[- \frac{π}{4}, \frac{π}{4}] is

Answered by Tinkutara last updated on 28/Jul/17

Using AM ⩾ GM

\frac{\sin^{3} x + \cos^{3} x + \cos^{3} x}{3} ⩾ \sin x \cos^{2} x

\sin^{3} x + \cos^{3} x + \cos^{3} x ⩾ 3 \sin x \cos^{2} x

∴ \sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \frac{π}{4}

Hence 1 solution in [- \frac{π}{4}, \frac{π}{4}] .

Answered by ajfour last updated on 29/Jul/17

\sin^{3} x - \sin xcos^{2} x - 2 \sin xcos^{2} x

+ 2 \cos^{3} x = 0

\sin x (\sin^{2} x - \cos^{2} x) - 2 \cos^{2} x (\sin x - \cos x) = 0

(\sin x - \cos x) (\sin^{2} x + \sin xcos x - 2 \cos^{2} x) = 0

(\sin x - \cos x) (\sin^{2} x - \sin xcos x

+ 2 \sin xcos x - 2 \cos^{2} x) = 0

\Rightarrow (\sin x - \cos x) (\sin x - \cos x)

\times (\sin x - 2 \cos x) = 0

{(\sin x - \cos x)}^{2} (\sin x - 2 \cos x) = 0

\Rightarrow \sin x = \cos x or

\sin x = 2 \cos x

I n [- \frac{π}{4}, \frac{π}{4}] only solution is x = \frac{π}{4} .

Commented by Tinkutara last updated on 29/Jul/17

Thanks Sir! Also a nice method .