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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

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

Switch sides

sin (x) = \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 (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

sin (3 x) - sin (x) = - 2 sin (2 x) sin (x)

x, 0.05 cos (x) = - sin (x)

cos (x^{\circ}) = \frac{20.26}{26.4473}

cos (x) = \frac{- \frac{4}{3} 3}{\frac{8}{3} 3}

cos (θ) = \frac{3}{10}

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