What is the general solution for 2cos(x)-1=-1 ?

The general solution for 2cos(x)-1=-1 is x= pi/2+2pin,x=(3pi)/2+2pin

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2cos(x)-1=-1

2 cos (x) - 1 = - 1

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (x) - 1 = - 1

Add

1

to both sides

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

Simplify

2 cos (x) = 0

Divide both sides by

2

2 cos (x) = 0

Divide both sides by

2

\frac{2 cos ( x )}{2} = \frac{0}{2}

Simplify

cos (x) = 0

cos (x) = 0

General solutions for

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

- sec (x) = csc (3.45)

cos (2 x) = \frac{8}{17}

tan (x) (csc (x) + 2) = 0

4 cos^{2} (x) = 5 + 4 sin (x), 0 \leq x \leq 2 π

sin (π + θ) = sin (θ)

What is the general solution for 2cos(x)-1=-1 ?
The general solution for 2cos(x)-1=-1 is x= pi/2+2pin,x=(3pi)/2+2pin