What is the general solution for 2cos(x)-2=0 ?

The general solution for 2cos(x)-2=0 is x=2pin

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2cos(x)-2=0

2 cos (x) - 2 = 0

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (x) - 2 = 0

Move

2

to the right side

2 cos (x) - 2 = 0

Add

2

to both sides

2 cos (x) - 2 + 2 = 0 + 2

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

4 sin^{2} (θ) - 13 sin (θ) + 9 = 0

23 cos^{2} (x) = sin (x)

4 cos^{4} (θ) - 5 cos^{2} (θ) + 1 = 0

x, cos (x) = sin (2 x)

sin (2 t) - sin (t) = 0

What is the general solution for 2cos(x)-2=0 ?
The general solution for 2cos(x)-2=0 is x=2pin