What is the general solution for solvefor x,cos(kx)=-1 ?

The general solution for solvefor x,cos(kx)=-1 is x= pi/k+(2pin)/k ;\quad k\ne 0

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solvefor x,cos(kx)=-1

solve for

x, cos (k x) = - 1

x = \frac{π}{k} + \frac{2 πn}{k}; k \neq = 0

Solution steps

cos (k x) = - 1

General solutions for

cos (k x) = - 1

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}

k x = π + 2 πn

k x = π + 2 πn

Solve

k x = π + 2 πn : x = \frac{π}{k} + \frac{2 πn}{k}; k \neq = 0

k x = π + 2 πn

Divide both sides by

k; k \neq = 0

k x = π + 2 πn

Divide both sides by

k; k \neq = 0

\frac{k x}{k} = \frac{π}{k} + \frac{2 πn}{k}; k \neq = 0

Simplify

x = \frac{π}{k} + \frac{2 πn}{k}; k \neq = 0

x = \frac{π}{k} + \frac{2 πn}{k}; k \neq = 0

x = \frac{π}{k} + \frac{2 πn}{k}; k \neq = 0

sec^{2} (x) = \frac{3}{4}

sin (x) = \frac{4.4 sin ( 11 9 ^{\circ} )}{12.3}

- sec^{2} (x) = 2 tan (x)

3 cot^{2} (x) - 1 = 0

sin (x) \cdot cos (x) = 1

What is the general solution for solvefor x,cos(kx)=-1 ?
The general solution for solvefor x,cos(kx)=-1 is x= pi/k+(2pin)/k ;\quad k\ne 0