Question

Solve the equation $si{n}^{3}xcos3x+co{s}^{3}xsin3x+\frac{3}{8}=0$

• A. $x=\frac{n\pi }{4}-(-1{)}^{n+1}\frac{\pi }{6};nϵz$
• B. $x=\frac{n\pi }{4}\pm \frac{\pi }{6};nϵz$
• C. $x=\frac{n\pi }{4}+(-1{)}^{n+1}\frac{\pi }{6};nϵz$
• D. $x=\frac{n\pi }{4}+(-1{)}^{n+1}\frac{\pi }{3};nϵz$

Answer 1

Answer: (C)

$x=\frac{n\pi }{4}+(-1{)}^{n+1}\frac{\pi }{6};nϵz$

${\sin }^{3}x\cos 3x+{\cos }^{3}x\sin 3x+\frac{3}{8}=0$
$\Rightarrow {\sin }^{3}x(4{\cos }^{3}x-3\cos x)+{\cos }^{3}x(3\sin x-4{\sin }^{3}x)+\frac{3}{8}=0$
$\Rightarrow 4{\sin }^{3}x{\cos }^{3}x-3\cos x{\sin }^{3}x+3\sin x{\cos }^{3}x-4{\sin }^{3}x{\cos }^{3}x+\frac{3}{8}=0$
$\Rightarrow 3\cos x\sin x({\cos }^{2}x-{\sin }^{2}x)+\frac{3}{8}=0$
$\Rightarrow \frac{3}{2}\sin 2x\cos 2x+\frac{3}{8}=0\Rightarrow \frac{3}{4}\sin 4x+\frac{3}{8}=0$
$\Rightarrow \sin 4x=-\frac{1}{2}\Rightarrow 4x=n\pi +{(-1)}^n(-\frac{\pi }{6})$