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Question Number 53210 by Tawa1 last updated on 19/Jan/19

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19

$$r_1^{{}^{\prime }}=r_1-r_3$$$$r_2^{,}=r_2-r_3$$$$\mid 10-1\mid$$$$\mid 01-1\mid$$$$\mid {sin}^2\theta {cos}^2\theta 1+4sin2\theta \mid$$$$=1(1+4sin2\theta +{cos}^2\theta )+(-1)(0-{sin}^2\theta )$$$$=1+4sin2\theta +1$$$$=2+4sin2\theta$$$$given4sin2\theta +2=0$$$$sin2\theta =-\frac{1}{2}=sin(\pi +\frac{\pi }{6})$$$$2\theta =\frac{7\pi }{6}\theta =\frac{7\pi }{12}\leftarrow firstsolution$$$$sin2\theta =-\frac{1}{2}=sin(2\pi -\frac{\pi }{6})$$$$2\theta =\frac{11\pi }{6}\theta =\frac{11\pi }{12}\leftarrow secondsolution$$

Commented by Tawa1 last updated on 19/Jan/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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