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

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

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-cos(x)=1

- cos (x) = 1

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

- cos (x) = 1

Divide both sides by

- 1

- cos (x) = 1

Divide both sides by

- 1

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

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 = π + 2 πn

x = π + 2 πn

cot (x) sin (x) - sin (x) = 0

1.16 cos (θ) + 0.532 cos (θ) - 0.557 = 0

arcsin (6 x) + arcsin (63 x) = - \frac{π}{2}

cos (x) = \frac{1}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}

- 44 sin (2 x) + 42 cos (2 x) = 0

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