What is the general solution for 2cos(x)+1=3,0<= x<= 2pi ?

The general solution for 2cos(x)+1=3,0<= x<= 2pi is x=0,x=2pi

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2cos(x)+1=3,0<= x<= 2pi

2 cos (x) + 1 = 3, 0 \leq x \leq 2 π

x = 0, x = 2 π

Degrees

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

Solution steps

2 cos (x) + 1 = 3, 0 \leq x \leq 2 π

Move

1

to the right side

2 cos (x) + 1 = 3

Subtract

1

from both sides

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

Simplify

2 cos (x) = 2

2 cos (x) = 2

Divide both sides by

2

2 cos (x) = 2

Divide both sides by

2

\frac{2 cos ( x )}{2} = \frac{2}{2}

Simplify

cos (x) = 1

cos (x) = 1

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

1 - 2 sin (2 x) = 0

2 sin (x) + \frac{1}{sin ( x )} = 0

2 cos^{2} (θ) = 2 + sin (θ)

cos (x) - cos^{3} (x) = 0

cos (x) = - cos (x)

What is the general solution for 2cos(x)+1=3,0<= x<= 2pi ?
The general solution for 2cos(x)+1=3,0<= x<= 2pi is x=0,x=2pi