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

The general solution for -1/(sqrt(2))=cos(x) is x=(3pi)/4+2pin,x=(5pi)/4+2pin

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

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

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Degrees

x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 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{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

3 sin (x) = sin (3 x)

tan (x) + 3 = 1 + 3 cot (x)

tan (u) - cot (u) = 2

sin (x) sin (x) = 0

15 = arctan (\frac{0.375}{2 x})

What is the general solution for -1/(sqrt(2))=cos(x) ?
The general solution for -1/(sqrt(2))=cos(x) is x=(3pi)/4+2pin,x=(5pi)/4+2pin