Question

Find the general solution of $1+{\sin }^{3}x+{\cos }^{3}x=\frac{3}{2}\sin 2x$

• A. $x=2n\pi \pm \frac{3\pi }{4},nϵz$
• B. $x=n\pi \pm \frac{3\pi }{4}+\frac{\pi }{4},nϵz$
• C. $x=2n\pi \pm \frac{3\pi }{4}+\frac{\pi }{4},nϵz$
• D. $x=2n\pi \pm \frac{\pi }{4}+\frac{\pi }{4},nϵz$

Answer 1

Answer: (C)

$x=2n\pi \pm \frac{3\pi }{4}+\frac{\pi }{4},nϵz$

$1+{\sin }^{3}x+{\cos }^{3}x=3\sin \left(x\right).\cos \left(x\right).1$
Hence
$a^3+b^3+c^3=3abc$ ...(i)
Now
$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc$ ...(ii)
Comparing i and ii, we get
$(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0$
Or
$a+b+c=0$
That implies
$1+\sin \left(x\right)+\cos \left(x\right)=0$
Or
$2{\cos }^2\left(\frac{x}{2}\right)+2\sin \frac{x}{2}.\cos \frac{x}{2}=0$

$2\cos \left(\frac{x}{2}\right)[\sin \frac{x}{2}+\cos \frac{x}{2}]=0$
Hence
$\cos \left(\frac{x}{2}\right)=0$ and $\sin \frac{x}{2}+\cos \frac{x}{2}=0$

$\frac{x}{2}=\frac{(2n+1)\pi }{2}$

$x=(2n+1)\pi$ ...(b) and
$\sin \frac{x}{2}+\cos \frac{x}{2}=0$

$\sqrt{2}\cos (\frac{x}{2}-\frac{\pi }{4})=0$

$\frac{x}{2}-\frac{\pi }{4}=\frac{(2n+1)\pi }{2}$

$x=(2n+1)\pi +\frac{\pi }{2}$. ..(a)
Hence from a and b
$x=2n\pi \pm \frac{3\pi }{4}+\frac{\pi }{4}$.