The number of θ ∈ (0, 4π) for which the system of linear equations...

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{cases}3\sin 3\theta & - 1 & 1 \\ 3\cos 2\theta & 4 & 3 \\ 6 & 7 & 7\end{cases}\right| = 0

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

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

$⇒ 2 - 2\cos 2\theta = \sin 3\theta$

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

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

$⇒ sin\theta = 0$ or \left(4sin\theta - 3 + 4\sin^{2}\theta\right) = 0$</p><p>Now$sin\theta = 0 ⇒ \theta \in \left\{\pi, 2\pi, 3\pi\right\}$</p><p>And \left(4sin\theta - 3 + 4\sin^{2}\theta\right) = 0$

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

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

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

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

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

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