The system of equations: 2x cos^2 theta + y s | vedclass

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(B) The given system of equations can be written as:$1) 2x \cos^2 	heta + y \sin 2	heta = 2 \sin 	heta$$2) x \sin 2	heta + 2y \sin^2 	heta = -2 \

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(B) The given system of equations can be written as:$1) 2x \cos^2 	heta + y \sin 2	heta = 2 \sin 	heta$$2) x \sin 2	heta + 2y \sin^2 	heta = -2 \cos 	heta$$3) x \sin 	heta - y \cos 	heta = 0$From equation $(3)$,we have $x \sin 	heta = y \cos 	heta$,which implies $y = x 	an 	heta$ (assuming $\cos 	heta 
eq 0$).Substituting this into equation $(1)$:$2x \cos^2 	heta + (x 	an 	heta)(2 \sin 	heta \cos 	heta) = 2 \sin 	heta$$2x \cos^2 	heta + 2x \sin^2 	heta = 2 \sin 	heta$$2x(\cos^2 	heta + \sin^2 	heta) = 2 \sin 	heta \Rightarrow 2x = 2 \sin 	heta \Rightarrow x = \sin 	heta$.Then $y = \sin 	heta \cdot 	an 	heta = \frac{\sin^2 	heta}{\cos 	heta}$.Substituting these values into equation $(2)$:$(\sin 	heta)(2 \sin 	heta \cos 	heta) + 2(\frac{\sin^2 	heta}{\cos 	heta})(\sin^2 	heta) = -2 \cos 	heta$$2 \sin^2 	heta \cos 	heta + \frac{2 \sin^4 	heta}{\cos 	heta} = -2 \cos 	heta$Multiplying by $\cos 	heta$:$2 \sin^2 	heta \cos^2 	heta + 2 \sin^4 	heta = -2 \cos^2 	heta$$2 \sin^2 	heta (\cos^2 	heta + \sin^2 	heta) = -2 \cos^2 	heta$$2 \sin^2 	heta = -2 \cos^2 	heta \Rightarrow \sin^2 	heta + \cos^2 	heta = 0 \Rightarrow 1 = 0$.This is a contradiction. Thus,the system of equations has no solution.

The system of equations: $2x \cos^2 \theta + y \sin 2\theta - 2 \sin \theta = 0$,$x \sin 2\theta + 2y \sin^2 \theta = -2 \cos \theta$,and $x \sin \theta - y \cos \theta = 0$,for all values of $\theta$,can:

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