The number of θ ∈ 0 , 4 π for which the system of linear equations 3 sin 3 θ x - y + z…

The number of $\theta \in \left(0,4\pi \right)$ for which the system of linear equations

$3\left(\sin 3\theta \right)x-y+z=2$

$3\left(\cos 2\theta \right)x+4y+3z=3$

$6x+7y+7z=9$ has no solution is :

We know that for the system of equation has no solution,

$D=\left|\begin{array}{ccc}3\sin 3\theta & -1& 1\\ 3\cos 2\theta & 4& 3\\ 6& 7& 7\end{array}\right|=0$

$\Rightarrow 21\sin 3\theta +42\cos 2\theta -42=0$

$\Rightarrow \sin 3\theta +2\cos 2\theta -2=0$

$\Rightarrow 2-2\cos 2\theta =\sin 3\theta$

$\Rightarrow 4{\sin }^2\theta =3\sin \theta -4{\sin }^3\theta$

$\Rightarrow \left(\sin \theta \right)\left(4\sin \theta -3+4{\sin }^2\theta \right)=0$

$\Rightarrow \sin \theta =0$ or $\left(4\sin \theta -3+4{\sin }^2\theta \right)=0$

Now $\sin \theta =0\Rightarrow \theta \in \left\{\mathrm{\pi },2\mathrm{\pi },3\mathrm{\pi }\right\}$

And $\left(4\sin \theta -3+4{\sin }^2\theta \right)=0$

 $\Rightarrow 4\sin \theta +4{\sin }^2\theta +1=4$

$\Rightarrow {\left(2\sin \theta +1\right)}^2=4$

$\Rightarrow \left(2\sin \theta +1\right)=\pm 2$

$\Rightarrow \sin \theta =\frac{1}{2}$ (ignoring negative sign as it will not lie in range of $\sin \theta$)

$\theta \in \left\{\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6},\frac{13\mathrm{\pi }}{6},\frac{17\mathrm{\pi }}{6}\right\}$

So, total number of solution is $7$ in $\left(0,4\pi \right)$