Question

The general solution satisfying the trigonometric equations
$4{\sin }^{3}x+1=\sin x(1+4\sin x)$ and
$\cos 3x=3–4{\cos }^{2}x$ is given by

• A. $x=n\pi \pm \frac{\pi }{6}$, only $n\in Z$
• B. $x=n\pi +(-1{)}^n\frac{\pi }{3}$ only, $n\in Z$
• C. $x=2n\pi \pm \frac{\pi }{6}$ only $n\in Z$
• D. $x=n\pi +(-{)}^n\frac{\pi }{6}$, $n\in Z$

Answer 1

The correct option is A $x=n\pi \pm \frac{\pi }{6}$, only $n\in Z$

$4{\sin }^{3}x-4{\sin }^{2}x-\sin x+1=0$
$(4{\sin }^{2}x-1)(\sin x-1)=0$
$\Rightarrow \sin x=\pm \frac{1}{2},\sin x=1$ ... (i)
$4{\cos }^{3}x-3\cos x=3-4{\cos }^{2}x$
$\Rightarrow 4{\cos }^{3}x+4{\cos }^{2}x-3\cos x-3=0$
$(4{\cos }^{2}x-3)(\cos x+1)=0$
$\cos x=\pm \frac{\sqrt{3}}{2},\cos x=-1$ ... (ii)
From (i) and (ii)
$x=n\pi \pm \frac{\pi }{6}$