If k is an integer which satisfies 2sin^2 θ+10cos^2 (θ/2)=7−2k, then k∈? A.{0,1,2,3} B.{−1,0,1,2,3} C.{(−2,−1,0,1,2,3} D.{−3,−2,−1,0}


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Question Number 117791 by ZiYangLee last updated on 13/Oct/20
$If k is an integer which satisfies 2sin^2 θ+10cos^2 (θ/2)=7−2k, then k∈? A.{0,1,2,3} B.{−1,0,1,2,3} C.{(−2,−1,0,1,2,3} D.{−3,−2,−1,0}$
$If k is an integer which satisfies$ ${2 \sin}^{2} θ + {10 \cos}^{2} \frac{θ}{2} = 7 - 2 k, then k \in ?$ $A . {0, 1, 2, 3} B . {- 1, 0, 1, 2, 3}$ $C . {(- 2, - 1, 0, 1, 2, 3} D . {- 3, - 2, - 1, 0}$
	Answered by floor(10²Eta[1]) last updated on 13/Oct/20

	$2 (1 - \cos^{2} θ) + 10 (\frac{1 + \cos θ}{2}) = 7 - 2 k$ $2 - {2 \cos}^{2} θ + 5 + 5 \cos θ = 7 - 2 k$ ${2 \cos}^{2} θ - 5 \cos θ - 2 k = 0$ $\cos θ = \frac{5 \pm \sqrt{25 + 16 k}}{4}$ $- 1 ⩽ \frac{5 \pm \sqrt{25 + 16 k}}{4} ⩽ 1$ $- 9 ⩽ \pm \sqrt{25 + 16 k} ⩽ - 1$ $\Rightarrow - 9 ⩽ - \sqrt{25 + 16 k} ⩽ - 1$ $9 ⩾ \sqrt{25 + 16 k} ⩾ 1$ $81 ⩾ 25 + 16 k ⩾ 1$ $3.5 ⩾ k ⩾ - 1.5$ $Ans : B$
	Answered by TANMAY PANACEA last updated on 13/Oct/20

	${c o s}^{2} \frac{θ}{2} = a$ $2 {(2 s i n \frac{θ}{2} c o s \frac{θ}{2})}^{2} + 10 {c o s}^{2} \frac{θ}{2}$ $8 (1 - {c o s}^{2} \frac{θ}{2}) {c o s}^{2} \frac{θ}{2} + 10 {c o s}^{2} \frac{θ}{2}$ $8 (1 - a) a + 10 a$ $8 a - 8 a^{2} + 10 a$ $18 a - 8 a^{2}$ $- 8 a^{2} + 18 a$ $- 8 (a^{2} - \frac{9 a}{4})$ $- 8 (a^{2} - 2 \times a \times \frac{9}{8} + \frac{81}{64} - \frac{81}{64})$ $8 \times \frac{81}{64} - 8 {(a - \frac{9}{8})}^{2}$ $\frac{81}{8} - 8 {(a - \frac{9}{8})}^{2}$ $1 ⩾ c o s \frac{θ}{2} ⩾ - 1 b u t 1 ⩾ {c o s}^{2} \frac{θ}{2} ⩾ 0$ $1 ⩾ a ⩾ 0$ $\frac{81}{8} - 8 {(0 - \frac{9}{8})}^{2} = 0$ $\frac{81}{8} - 8 {(1 - \frac{9}{8})}^{2} = \frac{81}{8} - 8 \times \frac{1}{64} = \frac{81 - 1}{8} = 10$ $n o w 10 ⩾ 7 - 2 k ⩾ 0$ $3 ⩾ - 2 k ⩾ - 7$ $- 3 ⩽ 2 k ⩽ 7$ $\frac{- 3}{2} ⩽ k ⩽ \frac{7}{2}$ $- 1.5 ⩽ k ⩽ 3.5$
	Answered by 1549442205PVT last updated on 14/Oct/20
	$2sin^2 θ+10cos^2 (θ/2)=7−2k (1) ⇔2(sin(θ/2)cos(θ/2))^2 +10cos^2 (θ/2)=7−2k ⇔2sin^2 (θ/2)cos^2 (θ/2)+10cos^2 (θ/2)=7−k •If cos^2 x=0 then sin^2 x=1⇒(1)⇔k=2.5∉Z •If cos^2 x≠0 then divide both sides of equation(1)by cos^2 x we get (1)⇔2tan^2 (θ/2)+10=(7−2k)(1+tan^2 (θ/2)) ⇔(5−2k)tan^2 (θ/2)=3+2k⇒k≠2.5 ⇔tan^2 (θ/2)=((3+2k)/(5−2k))≥0⇔−1.5≤k<2.5 k∈Z⇒k∈{−1,0,1,2}⇒Choose B$
	${2 \sin}^{2} θ + {10 \cos}^{2} \frac{θ}{2} = 7 - 2 k (1)$ $\Leftrightarrow 2 {(\sin \frac{θ}{2} \cos \frac{θ}{2})}^{2} + {10 \cos}^{2} \frac{θ}{2} = 7 - 2 k$ $\Leftrightarrow {2 \sin}^{2} \frac{θ}{2} \cos^{2} \frac{θ}{2} + {10 \cos}^{2} \frac{θ}{2} = 7 - k$ $∙ If \cos^{2} x = 0 then \sin^{2} x = 1 \Rightarrow (1) \Leftrightarrow k = 2.5 \notin Z$ $∙ If \cos^{2} x \neq 0 then divide both sides of$ $equation (1) by \cos^{2} x we get$ $(1) \Leftrightarrow {2 \tan}^{2} \frac{θ}{2} + 10 = (7 - 2 k) (1 + \tan^{2} \frac{θ}{2})$ $\Leftrightarrow (5 - 2 k) \tan^{2} \frac{θ}{2} = 3 + 2 k \Rightarrow k \neq 2.5$ $\Leftrightarrow \tan^{2} \frac{θ}{2} = \frac{3 + 2 k}{5 - 2 k} ⩾ 0 \Leftrightarrow - 1.5 ⩽ k < 2.5$ $k \in Z \Rightarrow k \in {- 1, 0, 1, 2} \Rightarrow Choose B$
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