#### Explain Solution R D Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 52 Sub Question 9 maths Textbook Solution.

Answer: $\theta=n \pi+\frac{(-1)^{n} \pi}{6}, n \in z \text { or } n \pi$

Hint: Use determinant formula

Given:$\left|\begin{array}{ccc} 3 & -2 & \sin 3 \theta \\ -7 & 8 & \cos 2 \theta \\ -11 & 14 & 2 \end{array}\right|=0$

Solution:

\begin{aligned} &\text { Apply C }_{1} \rightarrow C_{1}+C_{2} \\ &\begin{array}{lcc} 1 & -2 & \sin 3 \theta \\ 1 & 8 & \cos 2 \theta \\ 3 & 14 & 2 \end{array} \mid=0 \\ &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \& R_{3} \rightarrow R_{3}-R_{1} \\ &\left|\begin{array}{ccc} 1 & -2 & \sin 3 \theta \\ 0 & 10 & \cos 2 \theta-\sin 3 \theta \\ 0 & 20 & 2-3 \sin 3 \theta \end{array}\right|=0 \end{aligned}

\begin{aligned} &\text { Expand from } \mathrm{C}_{1}\\ &1[10(2-3 \sin \theta)-20(\cos 2 \theta-\sin 3 \theta)]=0\\ &2-3 \sin 3 \theta-2(\cos 2 \theta-\sin 3 \theta)=0\\ &2-3 \sin 3 \theta-2 \cos 2 \theta+2 \sin 3 \theta=0\\ &2-3 \sin 3 \theta-2 \cos 2 \theta=0\\ &2=3 \sin \theta-4 \sin ^{3} \theta+2\left(1-2 \sin ^{2} \theta\right)\\ &2=3 \sin \theta-4 \sin ^{3} \theta+2-4 \sin ^{2} \theta\\ &0=\sin \theta-\left(3-4 \sin ^{2} \theta-4 \sin \theta\right)\\ &-4 \sin ^{2} \theta-4 \sin \theta+3=0\\ &4 \sin ^{2} \theta+4 \sin \theta-3=0\\ &\sin \theta=x\\ &4 x^{2}+4 x-3=0 \end{aligned}

\begin{aligned} &4 x^{2}+6 x-2 x-3=0 \\ &2 x(2 x+3)-1(2 x+3)=0 \\ &2 x+3=0,2 x-1=0,2 x=0 \\ &x=\frac{-3}{2}, x=\frac{1}{2}, x=0, \\ &\sin \theta=\frac{-3}{2}, \sin \theta=\frac{1}{2}, \sin \theta=0 \\ &\text { As } \sin \theta=0, \theta=n \pi \& \text { as } \sin \theta=\frac{1}{2}, \theta=n \pi+\frac{(-1)^{n} \pi}{6}, n \in z \end{aligned}