【数学I】例題4.1.16：三角比を含む方程式の解の個数2（One More）★★★★

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% 例題I4.1.16：三角比を含む方程式の解の個数2（One More）★★★★
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方程式$2\sin^2\theta-a\sin\theta+1=0(0^{\circ}\leqq\theta\leqq 180^{\circ})$を満たす$\theta$が異なる4個の解をもつような定数$a$の値の範囲を求めよ．

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% 例題I4.1.16の解答（One More）★★★★
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$\sin\theta=t$とおくと，与えられた方程式は，$2t^2-at+1=0$となり，この2次方程式の判別式を$D$とする．

$0^{\circ}\leqq\theta\leqq 180^{\circ}$のとき，$t$の値の範囲は$0\leqq t\leqq 1$であり，$0\leqq t<1$のとき，$\sin\theta=t$を満たす$\theta$の値は2個であり，$t=1$のとき$\sin\theta=1$を満たす$\theta$の値は1個である．

したがって，与えられた方程式を満たす$\theta$が$0^{\circ}\leqq\theta\leqq 180^{\circ}$の範囲で異なる4個の解をもつのは，$2t^2-at+1=0$が$0\leqq t<1$の範囲で異なる2つの実数解をもつときである．

ゆえに，求める条件は，$f(t)=2t^2-at+1$とすると，(i)$D>0$，(ii)軸が$0<t<1$の間にある，(iii)$f(0)\geqq 0,f(1)>0$である．

(i) $D=a^2-4\cdot 2\cdot 1=a^2-8$であり，$D>0$であるから，
\[
a^2-8>0
\]

したがって，$a<-2\sqrt{2},2\sqrt{2}<a$

(ii) 軸は直線$t=\frac{a}{4}$であるから，$0<\frac{a}{4}<1$より，$0<a<4$

(iii) $f(0)=1,f(1)=-a+3$であり，$f(0)\geqq 0$かつ$f(1)>0$より，$-a+3>0$

したがって，$a<3$

よって，(i)〜(iii)より，求める$a$の値の範囲は，

\[
2\sqrt{2}<a<3
\]

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% 問題I4.1.16：三角比を含む方程式の解の個数2（One More）★★★★
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方程式$2\cos^2\theta+a\cos\theta+1=0(0^{\circ}\leqq\theta\leqq 90^{\circ})$を満たす$\theta$が異なる2個の解をもつような定数$a$の値の範囲を求めよ．

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% 問題I4.1.16の解答（One More）★★★★
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$\cos\theta=t$とおくと，与えられた方程式は$2t^2+at+1=0$となり，この2次方程式の判別式を$D$とする．

$0^{\circ}\leqq\theta\leqq 90^{\circ}$のとき，$t$の値の範囲は$0\leqq t\leqq 1$であり，$0\leqq t\leqq 1$のとき，$\cos\theta=t$を満たす$\theta$の値は1個である．

したがって，与えられた方程式を満たす$\theta$が$0^{\circ}\leqq\theta\leqq 90^{\circ}$の範囲で異なる2個の解をもつのは，$2t^2+at+1=0$が$0\leqq t\leqq 1$の範囲で異なる2つの実数解をもつときである．

ゆえに，求める条件は，$f(t)=2t^2+at+1$とすると，(i)$D>0$，(ii)軸が$0<t<1$の範囲にある，(iii)$f(0)\geqq 0,f(1)\geqq 0$である．

(i) $D=a^2-4\cdot 2\cdot 1=a^2-8$であり，$D>0$であるから，

\[
a^2-8>0
\]

したがって，$a<-2\sqrt{2},2\sqrt{2}<a$

(ii) 軸は直線$t=-\frac{a}{4}$であるから，$0<-\frac{a}{4}<1$より，$-4<a<0$

(iii) $f(0)=1,f(1)=a+3$であり，$f(0)\geqq 0$かつ$f(1)\geqq 0$より，$a+3\geqq 0$

したがって，$-3\leqq a$

よって，(i)〜(iii)より，求める$a$の値の範囲は，

\[
-3\leqq a<-2\sqrt{2}
\]

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% 例題I4.1.16の側注：三角比を含む方程式の解の個数2（One More）★★★★
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% 側注1
(i)のように判別式$D$を用いる代わりに，「頂点の$y$座標が$0$より小さくなる」という条件を用いても同じ結果が得られる．

% 側注2
おき換えた文字の範囲に注意すること．

% 側注3
$t=1$は含まないので注意すること．

% 側注4
$f(0)\geqq 0$は常に成り立つ．

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% 問題I4.1.16の側注：三角比を含む方程式の解の個数2（One More）★★★★
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% 側注1
(i)のように判別式$D$を用いる代わりに，「頂点の$y$座標が$0$より小さくなる」という条件を用いても同じ結果が得られる．

% 側注2
おき換えた文字の範囲に注意すること．

% 側注3
$f(0)\geqq 0$は常に成り立つ．

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% 例題I4.1.16の考え方・補足など：三角比を含む方程式の解の個数2（One More）★★★★
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% 考え方
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解の存在範囲は，判別式（頂点），軸，端点に注目して考える．$\sin\theta=t$とおくと，$0\leqq t<1(0^{\circ}\leqq\theta\leqq 180^{\circ})$のとき，1つの$t$に対して$\theta$が2個存在することから，$t$が$0\leqq t<1$において異なる2つの実数解をもつとき，4個の解をもつことになることを利用する．