Question

the values of θ lying between θ = 0 & θ = π/2 and satisfying the equation
$\left|\begin{array}{ccc} \sf 1+sin^2\theta &\sf cos^2\theta &\sf 4sin\:4\theta\\\sf sin^2\theta &\sf 1+cos^2\theta &\sf 4sin\:4\theta\\\sf sin^2\theta &\sf cos^2\theta&\sf 1+4sin\:4\theta \end{array}\right|$​

Answer 1

$\sf 0 = \left|\begin{array}{ccc} \sf 1+sin^2\theta &\sf cos^2\theta &\sf 4sin\:4\theta\\\sf sin^2\theta &\sf 1+cos^2\theta &\sf 4sin\:4\theta\\\sf sin^2\theta &\sf cos^2\theta&\sf 1+4sin\:4\theta \end{array}\right|$

$= \left|\begin{array}{ccc} \sf 2 &\sf cos^2\theta &\sf 4sin\:4\theta\\\sf 2 &\sf 1+cos^2\theta &\sf 4sin\:4\theta\\\sf 2 &\sf cos^2\theta&\sf 1+4sin\:4\theta \end{array}\right|$

$= \left|\begin{array}{ccc} \sf 2&\sf cos^2\theta &\sf 4sin\:4\theta\\\sf 0&\sf 1&\sf 0\\\sf -1&\sf 0&\sf 1\end{array}\right|$

now, to find determinant:-

$\left|\begin{array}{ccc} \boxed{+} &\sf \boxed{-} & \boxed{+}\\\\\boxed{-}&\boxed{+}&\boxed{-}\\\\\boxed{+}&\boxed{-}&\boxed{+}\end{array}\right|$

so now,

$\sf = +2\left|\begin{array}{cc} \sf 1&\sf 0\\\sf 0&\sf 1 \end{array}\right| - cos^2\theta \left|\begin{array}{cc}\sf 0&\sf 0\\\sf -1&\sf 1 \end{array}\right| + 4\:sin\:4\theta \left|\begin{array}{cc} \sf 0&\sf 1\\\sf -1&\sf 0 \end{array}\right|$

$\sf = +2(1)(1)-(0)(0)-cos^2\theta(0)(1)-(-1)(0)+4\:sin\:4\theta(0)(0)-(-1)(1)$

$\sf = 2-cos^2\theta \times 0 + 4\:sin\:4\theta$

$\sf = 2+4\:sin\:4\theta$

$\sf \therefore sin\:4\theta = \dfrac{-1}{2}$

$\sf \implies 4\theta = \dfrac{7\pi}{6} , \dfrac{4\pi}{6}$

$\sf \implies \theta = \dfrac{7\pi}{24} \;\: or \:\: \dfrac{4\pi}{24}$

Answer 2

Answer:

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