What is the general solution for cos(5x)=-1,0<= x<2pi ?

The general solution for cos(5x)=-1,0<= x<2pi is x= pi/5 ,x=(3pi)/5 ,x=pi,x=(7pi)/5 ,x=(9pi)/5

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cos(5x)=-1,0<= x<2pi

cos (5 x) = - 1, 0 \leq x < 2 π

x = \frac{π}{5}, x = \frac{3 π}{5}, x = π, x = \frac{7 π}{5}, x = \frac{9 π}{5}

Degrees

x = 3 6^{\circ}, x = 10 8^{\circ}, x = 18 0^{\circ}, x = 25 2^{\circ}, x = 32 4^{\circ}

Solution steps

cos (5 x) = - 1, 0 \leq x < 2 π

General solutions for

cos (5 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}

5 x = π + 2 πn

5 x = π + 2 πn

Solve

5 x = π + 2 πn : x = \frac{π}{5} + \frac{2 πn}{5}

5 x = π + 2 πn

Divide both sides by

5

5 x = π + 2 πn

Divide both sides by

5

\frac{5 x}{5} = \frac{π}{5} + \frac{2 πn}{5}

Simplify

x = \frac{π}{5} + \frac{2 πn}{5}

x = \frac{π}{5} + \frac{2 πn}{5}

x = \frac{π}{5} + \frac{2 πn}{5}

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{5}, x = \frac{3 π}{5}, x = π, x = \frac{7 π}{5}, x = \frac{9 π}{5}

sinh (x) = - 1

3 cos^{2} (x) + 2 sin (x) - 3 = 0

cot (x) = - 2

cos (c) = \frac{15}{17}

sin (θ) = 0.3907

What is the general solution for cos(5x)=-1,0<= x<2pi ?
The general solution for cos(5x)=-1,0<= x<2pi is x= pi/5 ,x=(3pi)/5 ,x=pi,x=(7pi)/5 ,x=(9pi)/5