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人気のある三角関数 >

solvefor x,sin^2(x)=(9m-8)/2

解

解く

x, sin^{2} (x) = \frac{9 m - 8}{2}

解

x = arcsin (\frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (\frac{9 m - 8}{2}) + 2 πn

解答ステップ

sin^{2} (x) = \frac{9 m - 8}{2}

置換で解く

sin^{2} (x) = \frac{9 m - 8}{2}

仮定:

sin (x) = u

u^{2} = \frac{9 m - 8}{2}

u^{2} = \frac{9 m - 8}{2} : u = \frac{9 m - 8}{2}, u = - \frac{9 m - 8}{2}

u^{2} = \frac{9 m - 8}{2}

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

u = \frac{9 m - 8}{2}, u = - \frac{9 m - 8}{2}

代用を戻す

u = sin (x)

sin (x) = \frac{9 m - 8}{2}, sin (x) = - \frac{9 m - 8}{2}

sin (x) = \frac{9 m - 8}{2}, sin (x) = - \frac{9 m - 8}{2}

sin (x) = \frac{9 m - 8}{2} : x = arcsin (\frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (- \frac{9 m - 8}{2}) + 2 πn

sin (x) = \frac{9 m - 8}{2}

三角関数の逆数プロパティを適用する

sin (x) = \frac{9 m - 8}{2}

以下の一般解

sin (x) = \frac{9 m - 8}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (- \frac{9 m - 8}{2}) + 2 πn

x = arcsin (\frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (- \frac{9 m - 8}{2}) + 2 πn

sin (x) = - \frac{9 m - 8}{2} : x = arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (\frac{9 m - 8}{2}) + 2 πn

sin (x) = - \frac{9 m - 8}{2}

三角関数の逆数プロパティを適用する

sin (x) = - \frac{9 m - 8}{2}

以下の一般解

sin (x) = - \frac{9 m - 8}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (\frac{9 m - 8}{2}) + 2 πn

x = arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (\frac{9 m - 8}{2}) + 2 πn

すべての解を組み合わせる

x = arcsin (\frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (\frac{9 m - 8}{2}) + 2 πn

グラフ

人気の例

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cos (θ) = \frac{12}{15}

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tan^2(x)+cot^2(x)-2=0

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