English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

cos(2θ)= 3/5 ,90<= θ<= 180

Lösung

cos (2 θ) = \frac{3}{5}, 9 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = \frac{36 0 ^{\circ} - 0.92729 \dots}{2}

Radianten

θ = \frac{2 π - 0.92729 \dots}{2}

Schritte zur Lösung

cos (2 θ) = \frac{3}{5}, 9 0^{\circ} \leq θ \leq 18 0^{\circ}

Wende die Eigenschaften der Trigonometrie an

cos (2 θ) = \frac{3}{5}

Allgemeine Lösung für

cos (2 θ) = \frac{3}{5}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

2 θ = arccos (\frac{3}{5}) + 36 0^{\circ} n, 2 θ = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n

2 θ = arccos (\frac{3}{5}) + 36 0^{\circ} n, 2 θ = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n

Löse

2 θ = arccos (\frac{3}{5}) + 36 0^{\circ} n : θ = \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

2 θ = arccos (\frac{3}{5}) + 36 0^{\circ} n

Teile beide Seiten durch

2

2 θ = arccos (\frac{3}{5}) + 36 0^{\circ} n

Teile beide Seiten durch

2

\frac{2 θ}{2} = \frac{arccos ( \frac{3}{5} )}{2} + \frac{36 0 ^{\circ} n}{2}

Vereinfache

θ = \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

θ = \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

Löse

2 θ = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n : θ = 18 0^{\circ} - \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

2 θ = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n

Teile beide Seiten durch

2

2 θ = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n

Teile beide Seiten durch

2

\frac{2 θ}{2} = 18 0^{\circ} - \frac{arccos ( \frac{3}{5} )}{2} + \frac{36 0 ^{\circ} n}{2}

Vereinfache

θ = 18 0^{\circ} - \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

θ = 18 0^{\circ} - \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

θ = \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n, θ = 18 0^{\circ} - \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

Lösungen für den Bereich

9 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = \frac{36 0 ^{\circ} - arccos ( \frac{3}{5} )}{2}

Zeige Lösungen in Dezimalform

θ = \frac{36 0 ^{\circ} - 0.92729 \dots}{2}

sin^{2} (x) - 4 cos (x) - 4 = 0, 0 \leq x \leq 2 π

so l v e f or x, sin (x) = 0.5

so l v e f or x, z = arcsin (\frac{y}{x})

3 csc (A) - 6 = 0

15 cos^{2} (x) + sin (x) - 9 = 0