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

The general solution for cos(5x)=1,0<= x<= 2pi is x=0,x=(2pi)/5 ,x=(4pi)/5 ,x=(6pi)/5 ,x=(8pi)/5

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

cos (5 x) = 1, 0 \leq x \leq 2 π

x = 0, x = \frac{2 π}{5}, x = \frac{4 π}{5}, x = \frac{6 π}{5}, x = \frac{8 π}{5}

Degrees

x = 0^{\circ}, x = 7 2^{\circ}, x = 14 4^{\circ}, x = 21 6^{\circ}, x = 28 8^{\circ}

Solution steps

cos (5 x) = 1, 0 \leq x \leq 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 = 0 + 2 πn

5 x = 0 + 2 πn

Solve

5 x = 0 + 2 πn : x = \frac{2 πn}{5}

5 x = 0 + 2 πn

0 + 2 πn = 2 πn

5 x = 2 πn

Divide both sides by

5

5 x = 2 πn

Divide both sides by

5

\frac{5 x}{5} = \frac{2 πn}{5}

Simplify

x = \frac{2 πn}{5}

x = \frac{2 πn}{5}

x = \frac{2 πn}{5}

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = \frac{2 π}{5}, x = \frac{4 π}{5}, x = \frac{6 π}{5}, x = \frac{8 π}{5}

sec (2 θ) = 2, 0 < θ < \frac{π}{2}

2 cos (2 θ) + 1 = 0

sec (\frac{5 θ}{4}) = 2

sin^{2} (x) = 1 - cos (x)

3 sec (x) - 4 cos (x) = 0

What is the general solution for cos(5x)=1,0<= x<= 2pi ?
The general solution for cos(5x)=1,0<= x<= 2pi is x=0,x=(2pi)/5 ,x=(4pi)/5 ,x=(6pi)/5 ,x=(8pi)/5