What is the general solution for 2cos(x)+4=5 ?

The general solution for 2cos(x)+4=5 is x= pi/3+2pin,x=(5pi)/3+2pin

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2cos(x)+4=5

2 cos (x) + 4 = 5

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (x) + 4 = 5

Move

4

to the right side

2 cos (x) + 4 = 5

Subtract

4

from both sides

2 cos (x) + 4 - 4 = 5 - 4

Simplify

2 cos (x) = 1

2 cos (x) = 1

Divide both sides by

2

2 cos (x) = 1

Divide both sides by

2

\frac{2 cos ( x )}{2} = \frac{1}{2}

Simplify

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (2 x - \frac{π}{2}) = - 1

cot (x) = - \frac{1}{3}

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

tan (3 x) = 0

2 cos^{2} (θ) = 2 - cos (2 θ)

What is the general solution for 2cos(x)+4=5 ?
The general solution for 2cos(x)+4=5 is x= pi/3+2pin,x=(5pi)/3+2pin