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

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

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

cos (π + x) = \frac{1}{2}

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

Radians

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

Solution steps

cos (π + 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 (π + x) = \frac{2}{2}

General solutions for

cos (π + 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}

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

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

Solve

π + x = \frac{π}{4} + 2 πn : x = \frac{π}{4} + 2 πn - π

π + x = \frac{π}{4} + 2 πn

Move

π

to the right side

π + x = \frac{π}{4} + 2 πn

Subtract

π

from both sides

π + x - π = \frac{π}{4} + 2 πn - π

Simplify

x = \frac{π}{4} + 2 πn - π

x = \frac{π}{4} + 2 πn - π

Solve

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

π + x = \frac{7 π}{4} + 2 πn

Move

π

to the right side

π + x = \frac{7 π}{4} + 2 πn

Subtract

π

from both sides

π + x - π = \frac{7 π}{4} + 2 πn - π

Simplify

x = \frac{7 π}{4} + 2 πn - π

x = \frac{7 π}{4} + 2 πn - π

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

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