What is the general solution for sin(x)= 1/(sqrt(2)),0<= x<= 2pi ?

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

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sin(x)= 1/(sqrt(2)),0<= x<= 2pi

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}

Solution steps

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}

2 7^{2} = 5 6^{2} + 3 5^{2} - (2 \cdot 56 \cdot 35 \cdot cos (x))

x, sin^{2} (x) - sin (x) - 2 = 0

cos (x - \frac{2 π}{3}) = \frac{1}{2}

3 sin (θ) = 3

cos (2 θ) + sin (θ) = 0, 0 \leq θ < 2 π

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