What is the general solution for sin(a)= 2/(sqrt(5)) ?

The general solution for sin(a)= 2/(sqrt(5)) is a=1.10714…+2pin,a=pi-1.10714…+2pin

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sin(a)= 2/(sqrt(5))

sin (a) = \frac{2}{5}

a = 1.10714 \dots + 2 πn, a = π - 1.10714 \dots + 2 πn

Radians

a = 1.10714 \dots + 6.28318 \dots n, a = π - 1.10714 \dots + 6.28318 \dots n

Solution steps

sin (a) = \frac{2}{5}

Simplify

\frac{2}{5} : \frac{2 5}{5}

\frac{2}{5}

Multiply by the conjugate

\frac{5}{5}

= \frac{2 5}{5 5}

55 = 5

55

Apply radical rule:

a a = a

55 = 5

= 5

= \frac{2 5}{5}

sin (a) = \frac{2 5}{5}

Apply trig inverse properties

sin (a) = \frac{2 5}{5}

General solutions for

sin (a) = \frac{2 5}{5}

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

a = arcsin (\frac{2 5}{5}) + 2 πn, a = π - arcsin (\frac{2 5}{5}) + 2 πn

a = arcsin (\frac{2 5}{5}) + 2 πn, a = π - arcsin (\frac{2 5}{5}) + 2 πn

Show solutions in decimal form

a = 1.10714 \dots + 2 πn, a = π - 1.10714 \dots + 2 πn

2 cot^{2} (x) - 5 csc (x) - 1 = 0

3 sin (x) tan (x) + 3 tan (x) = 0

sin^{2} (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

cos (A) = \frac{1}{3}

100 = \frac{50}{sin ( \frac{x}{2} )}

What is the general solution for sin(a)= 2/(sqrt(5)) ?
The general solution for sin(a)= 2/(sqrt(5)) is a=1.10714…+2pin,a=pi-1.10714…+2pin