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

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

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

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

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

Degrees

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

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

Switch sides

cos (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}

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

General solutions for

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

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

2 sin (4 x) = sin (2 x)

csc (x) sec (x) = 22

u, x = sinh (u)

sin (θ) = \frac{tan ( θ )}{sin ( θ )}

5 tan^{2} (t) - tan (t) = 0

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