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

$$\mathrm{The}\mathrm{values}\mathrm{of}\theta \mathrm{lying}\mathrm{between}0\mathrm{and}$$$$\frac{\pi }{2}\mathrm{and}\mathrm{satisfying}\mathrm{the}\mathrm{equation}$$$$\left|\begin{array}{ccc}1+{\sin }^2\theta & {\cos }^2\theta & 4\sin 4\theta \\ {\sin }^2\theta & 1+{\cos }^2\theta & 4\sin 4\theta \\ {\sin }^2\theta & {\cos }^2\theta & 1+{\sin }^4\theta \end{array}\right|=0\mathrm{are}$$

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

$$\left|\begin{array}{ccc}1+{\sin }^2\theta +{\cos }^2\theta & {\cos }^2\theta & 4\sin 4\theta \\ {\sin }^2\theta +1+{\cos }^2\theta & 1+{\cos }^2\theta & 4\sin 4\theta \\ {\sin }^2\theta +{\cos }^2\theta & {\cos }^2\theta & 1+{\sin }^{}4\theta \end{array}\right|=0$$$$by{C}_1\to C_1+C_2$$$$\left|\begin{array}{ccc}2& {\cos }^2\theta & 4\sin 4\theta \\ 2& 1+{\cos }^2\theta & 4\sin 4\theta \\ 1& {\cos }^2\theta & 1+{\sin }^{}4\theta \end{array}\right|=0$$$$\left|\begin{array}{ccc}0& -1& 0\\ 2& 1+{\cos }^2\theta & 4\sin 4\theta \\ 1& {\cos }^2\theta & 1+{\sin }^{}4\theta \end{array}\right|=0$$$$by{R}_1\to R_1-R_2$$$$2(1+{\sin }^{}4\theta )-4\sin 4\theta =0$$$$1+\sin 4\theta -2\sin 4\theta =0$$$$1-\sin 4\theta =0$$$$\sin 4\theta =1$$$$4\theta =\frac{\pi }{2}$$$$\theta =\frac{\pi }{8}$$

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

$$1+\sin 4\theta -2\sin 4\theta =0$$$$1-\sin 4\theta =0$$$$\mathrm{please}\mathrm{check}$$

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

$$Than\chi$$

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

$$\mathrm{welcome}$$

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