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

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

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

cos (2 x) = 1, 0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

Degrees

x = 0^{\circ}, x = 18 0^{\circ}, x = 36 0^{\circ}

Solution steps

cos (2 x) = 1, 0 \leq x \leq 2 π

General solutions for

cos (2 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}

2 x = 0 + 2 πn

2 x = 0 + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

x = πn

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

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

2 sin (θ) - 1 = 0

sin (\frac{19 π}{12})

sin (36 0^{\circ})

sin (θ) = - \frac{1}{2}

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