【数学I】例題4.1.9：三角比を含む方程式2（One More）★★★

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% 例題I4.1.9：三角比を含む方程式2（One More）★★★
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次の等式を満たす$\theta$の値を求めよ．

(1) $2\cos^2\theta+9\sin\theta-6=0(0^{\circ}\leqq\theta\leqq 180^{\circ})$

(2) $\sin\theta\tan\theta=-\frac{3}{2}(90^{\circ}<\theta\leqq 180^{\circ})$

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% 例題I4.1.9の解答（One More）★★★
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(1) $2\cos^2\theta+9\sin\theta-6=0$より，$2(1-\sin^2\theta)+9\sin\theta-6=0$

したがって，$2\sin^2\theta-9\sin\theta+4=0\cdots(\mathrm{i})$

$\sin\theta=t$とおくと，$0^{\circ}\leqq\theta\leqq 180^{\circ}$より，$0\leqq t\leqq 1$

これを(i)に代入すると，$2t^2-9t+4=0$

ゆえに，$(2t-1)(t-4)=0$より，$t=\frac{1}{2},4$

$0\leqq t\leqq 1$より，$t=\frac{1}{2}$

すなわち，$\sin\theta=\frac{1}{2}$

よって，$0^{\circ}\leqq\theta\leqq 180^{\circ}$より，$\theta=30^\circ,150^\circ$

(2) $\sin\theta\tan\theta=-\frac{3}{2}$より，$\sin\theta\cdot\frac{\sin\theta}{\cos\theta}=-\frac{3}{2}$

整理すると，$2\sin^2\theta=-3\cos\theta$

したがって，$2(1-\cos^2\theta)=-3\cos\theta$より，$2\cos^2\theta-3\cos\theta-2=0\cdots(\mathrm{ii})$

$\cos\theta=t$とおくと，$90^{\circ}<\theta\leqq 180^{\circ}$より，$-1\leqq t<0$

これを(ii)に代入すると，$2t^2-3t-2=0$

ゆえに，$(2t+1)(t-2)=0$より，$t=-\frac{1}{2},2$

$-1\leqq t<0$より，$t=-\frac{1}{2}$

すなわち，$\cos\theta=-\frac{1}{2}$

よって，$90^{\circ}<\theta\leqq 180^{\circ}$より，$\theta=120^{\circ}$

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% 問題I4.1.9：三角比を含む方程式2（One More）★★★
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次の等式を満たす$\theta$の値を求めよ．

(1) $2\sin^2\theta-3\cos\theta=0(0^{\circ}\leqq\theta\leqq 180^{\circ})$

(2) $2\cos^2\theta+7\sin\theta-5=0(0^{\circ}\leqq\theta\leqq 180^{\circ})$

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% 問題I4.1.9の解答（One More）★★★
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(1) $2\sin^2\theta-3\cos\theta=0$より，$2(1-\cos^2\theta)-3\cos\theta=0$

したがって，$2\cos^2\theta+3\cos\theta-2=0\cdots(\mathrm{i})$

$\cos\theta=t$とおくと，$0^{\circ}\leqq\theta\leqq 180^{\circ}$より，$-1\leqq t\leqq 1$

これを(i)に代入すると，$2t^2+3t-2=0$

ゆえに，$(2t-1)(t+2)=0$より，$t=\frac{1}{2},-2$

$-1\leqq t\leqq 1$より，$t=\frac{1}{2}$

すなわち，$\cos\theta=\frac{1}{2}$

よって，$0^{\circ}\leqq\theta\leqq 180^{\circ}$より，$\theta=60^\circ$

(2) $2\cos^2\theta+7\sin\theta-5=0$より，$2(1-\sin^2\theta)+7\sin\theta-5=0$

したがって，$2\sin^2\theta-7\sin\theta+3=0\cdots(\mathrm{i})$

$\sin\theta=t$とおくと，$0^{\circ}\leqq\theta\leqq 180^{\circ}$より，$0\leqq t\leqq 1$

これを(i)に代入すると，$2t^2-7t+3=0$

ゆえに，$(2t-1)(t-3)=0$より，$t=\frac{1}{2},3$

$0\leqq t\leqq 1$より，$t=\frac{1}{2}$

すなわち，$\sin\theta=\frac{1}{2}$

よって，$0^{\circ}\leqq\theta\leqq 180^{\circ}$より，$\theta=30^\circ,150^\circ$

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% 例題I4.1.9の側注：三角比を含む方程式2（One More）★★★
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% 側注1
$\sin^2\theta+\cos^2\theta=1$

% 側注2
$0^{\circ}\leqq\theta\leqq 180^{\circ}$のとき，$0\leqq\sin\theta\leqq 1$

% 側注3
たすき掛けを用いる．

% 側注4
$\tan\theta=\frac{\sin\theta}{\cos\theta}$

% 側注5
$\sin^2\theta+\cos^2\theta=1$

% 側注6
$90^{\circ}<\theta\leqq 180^{\circ}$のとき，$-1\leqq\cos\theta<0$

% 側注7
たすき掛けを用いる．

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% 問題I4.1.9の側注：三角比を含む方程式2（One More）★★★
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% 側注1
$\sin^2\theta+\cos^2\theta=1$

% 側注2
$0^{\circ}\leqq\theta\leqq 180^{\circ}$のとき，$-1\leqq\cos\theta\leqq 1$

% 側注3
たすき掛けを用いる．

% 側注4
$\sin^2\theta+\cos^2\theta=1$

% 側注5
$0^{\circ}\leqq\theta\leqq 180^{\circ}$のとき，$0\leqq\sin\theta\leqq 1$

% 側注6
たすき掛けを用いる．

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% 例題I4.1.9の考え方・補足など：三角比を含む方程式2（One More）★★★
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% 考え方
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$\sin^2\theta+\cos^2\theta=1,\tan\theta=\frac{\sin\theta}{\cos\theta}$などを利用して，三角比の種類を減らす．例えば，(1)は$\sin\theta,\cos\theta$の2種類の式であるから，種類を減らし，$\sin\theta$の1種類の式にする．

また，三角比を文字におき換えると，式が見やすくなり，扱いやすくなることがあるので，慣れないうちはおき換えることを推奨する．このとき，$\sin\theta,\cos\theta$の値の範囲から，おき換えた文字の範囲に注意すること．

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おき換えた文字の値の範囲に注意すること．