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

The general solution for cos(x)=1,0<= x<= 2pi is x=0,x=2pi

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

cos (x) = 1, 0 \leq x \leq 2 π

x = 0, x = 2 π

Degrees

x = 0^{\circ}, x = 36 0^{\circ}

Solution steps

cos (x) = 1, 0 \leq x \leq 2 π

General solutions for

cos (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}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = 2 π

x, cot (- x) cos (- x) + sin (- x) = \frac{- 1}{f}

tan^{2} (x) sin (x) = 3 sin (x)

cos (θ) - sin (2 θ) = 0

sin (7 x) = 0

cos (2 x) - 3 cos (x) = - 1

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