What is the general solution for sin(x)= 1/(sqrt(10)) ?

The general solution for sin(x)= 1/(sqrt(10)) is x=0.32175…+2pin,x=pi-0.32175…+2pin

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sin(x)= 1/(sqrt(10))

sin (x) = \frac{1}{10}

x = 0.32175 \dots + 2 πn, x = π - 0.32175 \dots + 2 πn

Radians

x = 0.32175 \dots + 6.28318 \dots n, x = π - 0.32175 \dots + 6.28318 \dots n

Solution steps

sin (x) = \frac{1}{10}

Simplify

\frac{1}{10} : \frac{10}{10}

\frac{1}{10}

Multiply by the conjugate

\frac{10}{10}

= \frac{1 \cdot 10}{10 10}

1 \cdot 10 = 10

1010 = 10

1010

Apply radical rule:

a a = a

1010 = 10

= 10

= \frac{10}{10}

sin (x) = \frac{10}{10}

Apply trig inverse properties

sin (x) = \frac{10}{10}

General solutions for

sin (x) = \frac{10}{10}

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

x = arcsin (\frac{10}{10}) + 2 πn, x = π - arcsin (\frac{10}{10}) + 2 πn

x = arcsin (\frac{10}{10}) + 2 πn, x = π - arcsin (\frac{10}{10}) + 2 πn

Show solutions in decimal form

x = 0.32175 \dots + 2 πn, x = π - 0.32175 \dots + 2 πn

(sin (θ) - 1) (sin (θ) + 1) = 0

5 sec (θ) - 6 = 0

sec (x) = - 1, - 2 π \leq x \leq 2 π

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

\frac{1}{cos ( 2 x )} + tan (2 x) = 3 cos (2 x)

What is the general solution for sin(x)= 1/(sqrt(10)) ?
The general solution for sin(x)= 1/(sqrt(10)) is x=0.32175…+2pin,x=pi-0.32175…+2pin