sin(θ)=(5n+1)/3-2n

Lösung

sin (θ) = \frac{5 n + 1}{3} - 2 n

θ = arcsin (\frac{5 n + 1}{3} - 2 n) + 2 πk, θ = π + arcsin (- \frac{5 n + 1}{3} + 2 n) + 2 πk

Schritte zur Lösung

sin (θ) = \frac{5 n + 1}{3} - 2 n

Wende die Eigenschaften der Trigonometrie an

sin (θ) = \frac{5 n + 1}{3} - 2 n

Allgemeine Lösung für

sin (θ) = \frac{5 n + 1}{3} - 2 n

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

θ = arcsin (\frac{5 n + 1}{3} - 2 n) + 2 πk, θ = π + arcsin (- \frac{5 n + 1}{3} + 2 n) + 2 πk

θ = arcsin (\frac{5 n + 1}{3} - 2 n) + 2 πk, θ = π + arcsin (- \frac{5 n + 1}{3} + 2 n) + 2 πk

cosh (x) - 1 = 0

arctan (θ) = arctan (\frac{92}{44})

10 cos (x) - 52 = 0

sin (2 x) = sin^{2} (2 x)

- 8 sin (2 θ + 4 5^{\circ}) = - 4, 0 \leq x < 36 0^{\circ}