Question

Solve the following equations.
$sin3x+si{n}^{3}x=\frac{3\sqrt{3}}{4}sin2x.$

Answer 1

Solution

$\to$ we are given that,

$sin3x+si{n}^{3}x=\frac{3\sqrt{3}}{4}sin2x$

$\to$ now, we know that $sin3\theta =3sin\theta -4si{n}^3\theta$

So, $3sinx-4si{n}^{3}x+si{n}^{3}x=\frac{3\sqrt{3}}{4}sin2x$

$\therefore 3sinx-\frac{3\sqrt{3}}{4}sin2x-3si{n}^{3}x=0$

$\therefore 3sinx-\frac{3\sqrt{3}}{4}\left(2sinxcosx\right)-3si{n}^{3}x=0$

$\therefore 3sinx-\frac{3\sqrt{3}}{2}sinxcosx-3si{n}^{3}x=0$

$\therefore 3sinx(1-\frac{\sqrt{3}}{2}cosx-si{n}^{2}x)=0$ (Take common)

$\therefore 3sinx(co{s}^{2}x-\frac{\sqrt{3}}{2}cosx)=0$ $(si{n}^{2}x.co{s}^{2}x=1)$

$\therefore 3sinxcosx(cosx-\frac{\sqrt{3}}{2})=0$

$\therefore 3/2\left(2sinxcosx\right)(cosx-\sqrt{3}/2)=0$

$\left(sin2x\right)(cosx-\sqrt{3}/1)=0$ $(\because sin2x=2sinxcosx)$

$sin2x=0$ or $cosx=\sqrt{3}/2.$

$2x=k\pi ,kϵz$ or $cos\alpha =\sqrt{3}/2\Rightarrow \alpha =\pi /6$

$x=k\pi /2,kϵz$ or $x=2k\pi \pm \pi /6,kϵz$

The required set of solution of x is

$\left\{k\pi /2/kϵ2\right\}\cup \{2k\pi \pm \pi /6/kϵz\}$