What is the general solution for sin(x)=-(sqrt(2))/2 ,-pi<= x<= pi ?

The general solution for sin(x)=-(sqrt(2))/2 ,-pi<= x<= pi is x=-(3pi)/4 ,x=-pi/4

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sin(x)=-(sqrt(2))/2 ,-pi<= x<= pi

sin (x) = - \frac{2}{2}, - π \leq x \leq π

x = - \frac{3 π}{4}, x = - \frac{π}{4}

Degrees

x = - 13 5^{\circ}, x = - 4 5^{\circ}

Solution steps

sin (x) = - \frac{2}{2}, - π \leq x \leq π

General solutions for

sin (x) = - \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}

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solutions for the range

- π \leq x \leq π

x = - \frac{3 π}{4}, x = - \frac{π}{4}

w, s (t) = A e^{- c t} cos (wt + b)

0 = 2 sin (t) + cos (2 t)

8 cos (θ) = - 4

2 sin (3 x + π) = - 1

t, f = cos (a t)

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