Question

[tex] \left |\begin{array}{ccc}1+sin²θ&cos²θ&4sin4θ\\sin²θ&1+cos²θ&4sin4θ \\ sin²θ&cos²θ &1+4sin4θ\end{array}\right | =0; 0 < θ < π/2,Solve it. [\tex]

Answer 1

Step-by-step explanation:

$\left |\begin{array}{ccc}1+sin^2\theta&cos^2\theta&4sin4\theta\\sin^2\theta&1+cos^2\theta&4sin4\theta \\sin^2\theta&cos^2\theta &1+4sin4\theta\end{array}\right | =0; 0 < θ < π/2$

$C_1 - > C_1 + C_2 + C_3$

$\left |\begin{array}{ccc}1+sin^2\theta+cos^2\theta+4sin4\theta&cos^2\theta&4sin4\theta\\1+sin^2\theta+cos^2\theta+4sin4\theta&1+cos^2\theta&4sin4\theta \\ 1+sin^2\theta+cos^2\theta+4sin4\theta&cos^2\theta &1+4sin4\theta\end{array}\right | =0; 0 < θ < π/2$

$\left |\begin{array}{ccc}2+4sin4\theta&cos^2\theta&4sin4\theta\\2+4sin4\theta&1+cos^2\theta&4sin4\theta \\ 2+4sin4\theta&cos^2\theta &1+4sin4\theta\end{array}\right | =0; 0 < θ < π/2$

Take common 2 +4 sin4θ from

$2+4sin4\theta\left |\begin{array}{ccc}1&cos^2\theta&4sin4\theta\\1&1+cos^2\theta&4sin4\theta \\ 1&cos^2\theta &1+4sin4\theta\end{array}\right | =0; 0 < θ < π/2$

$R_1 - > R_1 - R_3 \\ \\ R_2 - > R_2 - R_3$

$\left |\begin{array}{ccc}0&0&-1\\0&1&-1 \\ 1&cos^2\theta &1+4sin4\theta\end{array}\right | =0; 0 < θ < π/2$

Expand the determinant along R1

[tex](2+4sin4\theta)(1+cos^2\theta+4sin4\theta)=0\\\\(2+4sin4\theta)=0\\\\sin4\theta=\frac{-1}{2}\\\\4\theta=210\\\\\theta=52.5°\\\\(1+cos^2\theta+4sin4\theta)=0\\\\