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العربية

sin^2(x)= 1/2 ,0<= x<= 2pi

Lösung

sin^{2} (x) = \frac{1}{2}, 0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

Grad

x = 4 5^{\circ}, x = 13 5^{\circ}, x = 22 5^{\circ}, x = 31 5^{\circ}

Schritte zur Lösung

sin^{2} (x) = \frac{1}{2}, 0 \leq x \leq 2 π

Löse mit Substitution

sin^{2} (x) = \frac{1}{2}

Angenommen:

sin (x) = u

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2} : u = \frac{1}{2}, u = - \frac{1}{2}

u^{2} = \frac{1}{2}

Für

x^{2} = f (a)

sind die Lösungen

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

u = \frac{1}{2}, u = - \frac{1}{2}

Setze in

u = sin (x)

ein

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π

Allgemeine Lösung für

sin (x) = \frac{1}{2}

sin (x)

Periodizitätstabelle mit

2 πn

Zyklus:

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

Lösungen für den Bereich

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{5 π}{4}, x = \frac{7 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π

Allgemeine Lösung für

sin (x) = - \frac{1}{2}

sin (x)

Periodizitätstabelle mit

2 πn

Zyklus:

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Lösungen für den Bereich

0 \leq x \leq 2 π

x = \frac{5 π}{4}, x = \frac{7 π}{4}

Kombiniere alle Lösungen

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

sin (3 θ) = 1, 0 \leq θ < 2 π

cos (2 x) cos (x) - sin (2 x) sin (x) = \frac{3}{2}

sin (2 x) cos (x) = sin (x)

tan (\frac{1}{2} x) - 1 = 0

sin (3 x) = 0, 0 \leq x \leq 2 π