What is the general solution for -sqrt(2)sin(x)=1 ?

The general solution for -sqrt(2)sin(x)=1 is x=(5pi)/4+2pin,x=(7pi)/4+2pin

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-sqrt(2)sin(x)=1

- 2 sin (x) = 1

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

Degrees

x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

- 2 sin (x) = 1

Divide both sides by

- 2

- 2 sin (x) = 1

Divide both sides by

- 2

\frac{- 2 sin ( x )}{- 2} = \frac{1}{- 2}

Simplify

\frac{- 2 sin ( x )}{- 2} = \frac{1}{- 2}

Simplify

\frac{- 2 sin ( x )}{- 2} : sin (x)

\frac{- 2 sin ( x )}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

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

Cancel the common factor:

2

= sin (x)

Simplify

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

\frac{1}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{1}{2}

Rationalize

- \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}

= - \frac{2}{2}

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

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

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

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

- 5 sin (x) = - 2 cos^{2} (x) + 4, 0 \leq x \leq 2 π

6 = sin (\frac{π}{4} x) + 5

2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

t, 4.6 = 0.106 cos (4.36 t)

sin (x) = \frac{23}{24}

What is the general solution for -sqrt(2)sin(x)=1 ?
The general solution for -sqrt(2)sin(x)=1 is x=(5pi)/4+2pin,x=(7pi)/4+2pin