What is the general solution for sin(t)= 1/(sqrt(2)) ?

The general solution for sin(t)= 1/(sqrt(2)) is t= pi/4+2pin,t=(3pi)/4+2pin

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sin(t)= 1/(sqrt(2))

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

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

Solution steps

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

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

sin (t) = \frac{2}{2}

General solutions for

sin (t) = \frac{2}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

- 2 sin^{2} (θ) + 5 sin (θ) = 2

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

3 sin (x) - 2 = 0

4 cos^{2} (x) - 2 sin (x) + 1 = 0

tan (x) = \frac{3 3}{5}

What is the general solution for sin(t)= 1/(sqrt(2)) ?
The general solution for sin(t)= 1/(sqrt(2)) is t= pi/4+2pin,t=(3pi)/4+2pin