The-values-of-lying-between-0-and-pi-2-and-satisfying-the-equation-determinant-1-sin-2-cos-2-4-sin-4-sin-2-1-cos-2-4-sin-4-sin-2-cos-2-1-sin-4-

Question Number 108790 by saorey0202 last updated on 19/Aug/20

The values of θ lying between 0 and

\frac{π}{2} and satisfying the equation

| \begin{matrix} 1 + \sin^{2} θ & \cos^{2} θ & 4 \sin 4 θ \\ \sin^{2} θ & 1 + \cos^{2} θ & 4 \sin 4 θ \\ \sin^{2} θ & \cos^{2} θ & 1 + \sin^{4} θ \end{matrix} | = 0 are

Commented by $@y@m last updated on 19/Aug/20

I think there is typo in question. Please check.

Answered by $@y@m last updated on 19/Aug/20

| \begin{matrix} 1 + \sin^{2} θ + \cos^{2} θ & \cos^{2} θ & 4 \sin 4 θ \\ \sin^{2} θ + 1 + \cos^{2} θ & 1 + \cos^{2} θ & 4 \sin 4 θ \\ \sin^{2} θ + \cos^{2} θ & \cos^{2} θ & 1 + \sin^{} 4 θ \end{matrix} | = 0

b y C_{1} \to C_{1} + C_{2}

| \begin{matrix} 2 & \cos^{2} θ & 4 \sin 4 θ \\ 2 & 1 + \cos^{2} θ & 4 \sin 4 θ \\ 1 & \cos^{2} θ & 1 + \sin^{} 4 θ \end{matrix} | = 0

| \begin{matrix} 0 & - 1 & 0 \\ 2 & 1 + \cos^{2} θ & 4 \sin 4 θ \\ 1 & \cos^{2} θ & 1 + \sin^{} 4 θ \end{matrix} | = 0

b y R_{1} \to R_{1} - R_{2}

2 (1 + \sin^{} 4 θ) - 4 \sin 4 θ = 0

1 + \sin 4 θ - 2 \sin 4 θ = 0

1 - \sin 4 θ = 0

\sin 4 θ = 1

4 θ = \frac{π}{2}

θ = \frac{π}{8}

Commented by PRITHWISH SEN 2 last updated on 19/Aug/20

1 + \sin 4 θ - 2 \sin 4 θ = 0

1 - \sin 4 θ = 0

please check

Commented by $@y@m last updated on 19/Aug/20

T h a n χ

Commented by PRITHWISH SEN 2 last updated on 19/Aug/20

welcome