If sin2⁡θ=x2+y2+12x, then x must be

sin2⁡θ≤1⇒x2+y2+12x≤1⇒x2+y2−2x+1≤0⇒(x−1)2+y2≤0. It is possible, if x=1 and y=0

If sin2⁡θ=x2+y2+12x, then x must be

sin2⁡θ≤1⇒x2+y2+12x≤1⇒x2+y2−2x+1≤0⇒(x−1)2+y2≤0. It is possible, if x=1 and y=0

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$If \sin^{2} θ = \frac{x^{2} + y^{2} + 1}{2 x}, then x must be$

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$\sin^{2} θ \leq 1 \Rightarrow \frac{x^{2} + y^{2} + 1}{2 x} \leq 1 \Rightarrow x^{2} + y^{2} - 2 x + 1 \leq 0$
$\Rightarrow (x - 1)^{2} + y^{2} \leq 0 . It is possible, if x = 1 and y = 0$

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$If \sin^{2} θ = \frac{x^{2} + y^{2} + 1}{2 x}, then x must be$