JEE Main 2022 — Mathematics Algebra

The number of $\theta \in (0,4\pi )$θ∈(0,4π) for which the system of linear equations

$$3(\mathrm{sin}3\theta )x-y+z=2$$3(sin3θ)x−y+z=2

$$3(\mathrm{cos}2\theta )x+4y+3z=3$$3(cos2θ)x+4y+3z=3

$6x+7y+7z=9$6x+7y+7z=9 has no solution is :

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

$$D=|\begin{matrix}3\mathrm{sin}3\theta & -1 & 1 \\ 3\mathrm{cos}2\theta & 4 & 3 \\ 6 & 7 & 7\end{matrix}|=0$$D=∣3sin3θ3cos2θ6​−147​137​∣=0

$$\Rightarrow 21\mathrm{sin}3\theta +42\mathrm{cos}2\theta -42=0$$⇒21sin3θ+42cos2θ−42=0

$$\Rightarrow \mathrm{sin}3\theta +2\mathrm{cos}2\theta -2=0$$⇒sin3θ+2cos2θ−2=0

$$\Rightarrow 2-2\mathrm{cos}2\theta =\mathrm{sin}3\theta$$⇒2−2cos2θ=sin3θ

$$\Rightarrow 4{\mathrm{sin}}^{2}\theta =3\mathrm{sin}\theta -4{\mathrm{sin}}^{3}\theta$$⇒4sin2θ=3sinθ−4sin3θ

$$\Rightarrow (\mathrm{sin}\theta )(4\mathrm{sin}\theta -3+4{\mathrm{sin}}^{2}\theta )=0$$⇒(sinθ)(4sinθ−3+4sin2θ)=0

$\Rightarrow \mathrm{sin}\theta =0$⇒sinθ=0 or $(4\mathrm{sin}\theta -3+4{\mathrm{sin}}^{2}\theta )=0$(4sinθ−3+4sin2θ)=0

Now $\mathrm{sin}\theta =0\Rightarrow \theta \in {\pi ,2\pi ,3\pi }$sinθ=0⇒θ∈π,2π,3π

And $(4\mathrm{sin}\theta -3+4{\mathrm{sin}}^{2}\theta )=0$(4sinθ−3+4sin2θ)=0

$$\Rightarrow 4\mathrm{sin}\theta +4{\mathrm{sin}}^{2}\theta +1=4$$⇒4sinθ+4sin2θ+1=4

$$\Rightarrow {(2\mathrm{sin}\theta +1)}^{2}=4$$⇒(2sinθ+1)2=4

$$\Rightarrow (2\mathrm{sin}\theta +1)=\pm 2$$⇒(2sinθ+1)=±2

$\Rightarrow \mathrm{sin}\theta =\frac{1}{2}$⇒sinθ=21​ (ignoring negative sign as it will not lie in range of $\mathrm{sin}\theta$sinθ)

$$\theta \in {\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6}}$$θ∈6π​,65π​,613π​,617π​

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