Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

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

$$\mathrm{welcome} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com