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

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

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-2sin(t)=-sqrt(2)

- 2 sin (t) = - 2

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

Degrees

t = 4 5^{\circ} + 36 0^{\circ} n, t = 13 5^{\circ} + 36 0^{\circ} n

Solution steps

- 2 sin (t) = - 2

Divide both sides by

- 2

- 2 sin (t) = - 2

Divide both sides by

- 2

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

Simplify

sin (t) = \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

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

3 cos (3 t) = 0

2 cos (x) - 1 = 0, 0 \leq x \leq 2 π

18 cos (x) + 9 sin (2 x) = 0

sin (x) = \frac{5}{29}

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