What is the general solution for sqrt(2)sin(x)-1=0 ?

The general solution for sqrt(2)sin(x)-1=0 is x= pi/4+2pin,x=(3pi)/4+2pin

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sqrt(2)sin(x)-1=0

2 sin (x) - 1 = 0

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

Degrees

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

Solution steps

2 sin (x) - 1 = 0

Move

1

to the right side

2 sin (x) - 1 = 0

Add

1

to both sides

2 sin (x) - 1 + 1 = 0 + 1

Simplify

2 sin (x) = 1

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}

Cancel the common factor:

2

= sin (x)

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

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

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

cos (\frac{x}{2}) \frac{1}{2} = 0

2 sin (x) + csc (x) = 3

2 cos (x) + 1 = 0, 0 \leq x \leq 2 π

arctan (x) = \frac{π}{4}

What is the general solution for sqrt(2)sin(x)-1=0 ?
The general solution for sqrt(2)sin(x)-1=0 is x= pi/4+2pin,x=(3pi)/4+2pin