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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-

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Question Number 108790 by saorey0202 last updated on 19/Aug/20
The values of θ lying between 0 and (π/2) and satisfying the equation determinant (((1+sin^2 θ),( cos^2 θ),(4 sin 4θ)),(( sin^2 θ),(1+cos^2 θ),(4 sin 4θ)),(( sin^2 θ),( cos^2 θ),(1+sin^4 θ)))=0 are
$$\mathrm{The}\:\mathrm{values}\:\mathrm{of}\:\theta\:\mathrm{lying}\:\mathrm{between}\:\mathrm{0}\:\mathrm{and} \\ $$$$\frac{\pi}{\mathrm{2}}\:\mathrm{and}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\begin{vmatrix}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \theta}&{\:\:\mathrm{cos}^{\mathrm{2}} \theta}&{\mathrm{4}\:\mathrm{sin}\:\mathrm{4}\theta}\\{\:\:\:\mathrm{sin}^{\mathrm{2}} \theta}&{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \theta}&{\mathrm{4}\:\mathrm{sin}\:\mathrm{4}\theta}\\{\:\:\:\mathrm{sin}^{\mathrm{2}} \theta}&{\:\:\:\mathrm{cos}^{\mathrm{2}} \theta}&{\mathrm{1}+\mathrm{sin}^{\mathrm{4}} \theta}\end{vmatrix}=\mathrm{0}\:\:\mathrm{are} \\ $$
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
determinant (((1+sin^2 θ+cos^2 θ),( cos^2 θ),(4 sin 4θ)),(( sin^2 θ+1+cos^2 θ),(1+cos^2 θ),(4 sin 4θ)),(( sin^2 θ+cos^2 θ),( cos^2 θ),(1+sin^() 4θ)))=0 by C_1 →C_1 +C_2 determinant ((2,(cos^2 θ),(4 sin 4θ)),(2,(1+cos^2 θ),(4 sin 4θ)),(( 1),( cos^2 θ),(1+sin^() 4θ)))=0 determinant ((0,(−1),0),(2,(1+cos^2 θ),(4 sin 4θ)),(( 1),( cos^2 θ),(1+sin^() 4θ)))=0 by R_1 →R_1 −R_2 2(1+sin^() 4θ)−4sin 4θ=0 1+sin4θ−2sin 4θ=0 1−sin4θ=0 sin4θ=1 4θ=(π/2) θ=(π/8)
$$\begin{vmatrix}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta}&{\:\:\mathrm{cos}^{\mathrm{2}} \theta}&{\mathrm{4}\:\mathrm{sin}\:\mathrm{4}\theta}\\{\:\:\:\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \theta}&{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \theta}&{\mathrm{4}\:\mathrm{sin}\:\mathrm{4}\theta}\\{\:\:\:\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta}&{\:\:\:\mathrm{cos}^{\mathrm{2}} \theta}&{\mathrm{1}+\mathrm{sin}^{} \mathrm{4}\theta}\end{vmatrix}=\mathrm{0}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{by}\:{C}_{\mathrm{1}} \rightarrow{C}_{\mathrm{1}} +{C}_{\mathrm{2}} \\ $$$$\begin{vmatrix}{\mathrm{2}}&{\mathrm{cos}^{\mathrm{2}} \theta}&{\mathrm{4}\:\mathrm{sin}\:\mathrm{4}\theta}\\{\mathrm{2}}&{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \theta}&{\mathrm{4}\:\mathrm{sin}\:\mathrm{4}\theta}\\{\:\mathrm{1}}&{\:\:\:\mathrm{cos}^{\mathrm{2}} \theta}&{\mathrm{1}+\mathrm{sin}^{} \mathrm{4}\theta}\end{vmatrix}=\mathrm{0}\:\: \\ $$$$\begin{vmatrix}{\mathrm{0}}&{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{2}}&{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \theta}&{\mathrm{4}\:\mathrm{sin}\:\mathrm{4}\theta}\\{\:\mathrm{1}}&{\:\:\:\mathrm{cos}^{\mathrm{2}} \theta}&{\mathrm{1}+\mathrm{sin}^{} \mathrm{4}\theta}\end{vmatrix}=\mathrm{0}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{by}\:{R}_{\mathrm{1}} \rightarrow{R}_{\mathrm{1}} −{R}_{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{1}+\mathrm{sin}^{} \mathrm{4}\theta\right)−\mathrm{4sin}\:\mathrm{4}\theta=\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{sin4}\theta−\mathrm{2sin}\:\mathrm{4}\theta=\mathrm{0} \\ $$$$\mathrm{1}−\mathrm{sin4}\theta=\mathrm{0} \\ $$$$\mathrm{sin4}\theta=\mathrm{1} \\ $$$$\mathrm{4}\theta=\frac{\pi}{\mathrm{2}} \\ $$$$\theta=\frac{\pi}{\mathrm{8}} \\ $$
Commented by PRITHWISH SEN 2 last updated on 19/Aug/20
1+sin 4θ−2sin 4θ=0 1−sin 4θ=0 please check
$$\mathrm{1}+\mathrm{sin}\:\mathrm{4}\theta−\mathrm{2sin}\:\mathrm{4}\theta=\mathrm{0} \\ $$$$\mathrm{1}−\mathrm{sin}\:\mathrm{4}\theta=\mathrm{0} \\ $$$$\mathrm{please}\:\mathrm{check} \\ $$
Commented by $@y@m last updated on 19/Aug/20
Thanχ
$${Than}\chi \\ $$
Commented by PRITHWISH SEN 2 last updated on 19/Aug/20
welcome
$$\mathrm{welcome} \\ $$

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