What is the general solution for 2cos(t)=1,-pi<= t<= pi ?

The general solution for 2cos(t)=1,-pi<= t<= pi is t= pi/3 ,t=-pi/3

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2cos(t)=1,-pi<= t<= pi

2 cos (t) = 1, - π \leq t \leq π

t = \frac{π}{3}, t = - \frac{π}{3}

Degrees

t = 6 0^{\circ}, t = - 6 0^{\circ}

Solution steps

2 cos (t) = 1, - π \leq t \leq π

Divide both sides by

2

2 cos (t) = 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

cos (t) = \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}

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

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

Solutions for the range

- π \leq t \leq π

t = \frac{π}{3}, t = - \frac{π}{3}

tan (\frac{x}{3}) = 1, 0 \leq x \leq 2 π

4 sin (x) = sin (x) tan (x)

cos (x) = 0.707

4 cos^{2} (x) - 4 cos (x) - 3 = 0

cot (x) = \frac{4}{3}

What is the general solution for 2cos(t)=1,-pi<= t<= pi ?
The general solution for 2cos(t)=1,-pi<= t<= pi is t= pi/3 ,t=-pi/3