【数学I】例題4.1.15：三角比を含む方程式の解の個数1（One More）★★★★

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% 例題I4.1.15：三角比を含む方程式の解の個数1（One More）★★★★
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方程式$2\cos^2\theta+\sin\theta-a=0(0^{\circ}\leqq\theta\leqq 180^{\circ})$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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% 例題I4.1.15の解答（One More）★★★★
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$\sin\theta=t$とおくと，$\sin^2\theta+\cos^2\theta=1$より，与えられた方程式は，
\[
-2t^2+t+2-a=0\cdots(\mathrm{i})
\]

$0^{\circ}\leqq\theta\leqq 180^{\circ}$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，$t=1$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2個の解をもつのは，(i)が$0\leqq t<1$の範囲で解を1個もつときである．

(i)を整理すると，$a=-2t^2+t+2$

ゆえに，$a=-2(t-\frac{1}{4})^2+\frac{17}{8}\cdots(\mathrm{ii})$

(ii)の実数解の個数は，$y=a$と$y=-2(t-\frac{1}{4})^2+\frac{17}{8}$のグラフの共有点の個数と一致する．

よって，右の図より，$0\leqq t<1$の範囲で解を1個もつ$a$の範囲は，

\[
1<a<2,a=\frac{17}{8}
\]

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% 問題I4.1.15：三角比を含む方程式の解の個数1（One More）★★★★
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方程式$3\sin^2\theta+\cos\theta-a=0(0^{\circ}\leqq\theta\leqq 180^{\circ})$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I4.1.15の解答（One More）★★★★
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$\cos\theta=t$とおくと，$\sin^2\theta+\cos^2\theta=1$より，与えられた方程式は，
\[
-3t^2+t+3-a=0\cdots(\mathrm{i})
\]

$0^{\circ}\leqq\theta\leqq 180^{\circ}$のとき，$t$の値の範囲は$-1\leqq t\leqq 1$であり，$-1\leqq t\leqq 1$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$0^{\circ}\leqq\theta\leqq 180^{\circ}$の範囲で異なる2個の解をもつのは，(i)が$-1\leqq t\leqq 1$の範囲で解を2個もつときである．

(i)を整理すると，$a=-3t^2+t+3$

ゆえに，$a=-3(t-\frac{1}{6})^2+\frac{37}{12}\cdots(\mathrm{ii})$

(ii)の実数解の個数は，$y=a$と$y=-3(t-\frac{1}{6})^2+\frac{37}{12}$のグラフの共有点の個数と一致する．

よって，右の図より，$-1\leqq t\leqq 1$の範囲で解を2個もつ$a$の範囲は，

\[
1\leqq a<\frac{37}{12}
\]

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% 例題I4.1.15の側注：三角比を含む方程式の解の個数1（One More）★★★★
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% 側注1
定数を分離する．

% 側注2
$2\leqq a<\frac{17}{8}$のとき，共有点が2個存在し，それを満たす$\theta$は4個存在する．

% 側注3
$2(1-\sin^2\theta)+\sin\theta-a=0$より，$-2\sin^2\theta+\sin\theta+2-a=0$

% 側注4
$t=1$は含まないので注意すること．

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% 問題I4.1.15の側注：三角比を含む方程式の解の個数1（One More）★★★★
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% 側注1
定数を分離する．

% 側注2
$1\leqq a<\frac{37}{12}$のとき，共有点が2個存在し，それを満たす$\theta$は2個存在する．

% 側注3
$3(1-\cos^2\theta)+\cos\theta-a=0$より，$-3\cos^2\theta+\cos\theta+3-a=0$

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% 例題I4.1.15の考え方・補足など：三角比を含む方程式の解の個数1（One More）★★★★
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% 考え方
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先に，$\sin^2\theta+\cos^2\theta=1$を利用して，三角比の種類を減らす．次に，$\sin\theta=t$とおいて，定数を分離し，$y=a$と放物線$y=-2t^2+t+2$の共有点を考えればよい．このとき，$0^{\circ}\leqq\theta\leqq 180^{\circ}(\theta\neq 90^{\circ})$のとき，$\sin\theta=t(0\leqq t<1)$となる$\theta$は1つの$t$に対して2個存在し，$\theta=90^{\circ}$のとき，$\sin\theta=t=1$となる$\theta$は1つの$t$に対して1個存在することに注意すること．$0^{\circ}\leqq\theta\leqq 180^{\circ}(\theta\neq 90^{\circ})$のとき

1つの$t$に対して，$\theta$は2個存在する．

$\theta=90^{\circ}$のとき

1つの$t$に対して，$\theta$は1個存在する．