What is the general solution for 2cos(t)-1=0,0<,=t<= 2pi ?

The general solution for 2cos(t)-1=0,0<,=t<= 2pi is No Solution for t\in\mathbb{R}

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2cos(t)-1=0,0<,=t<= 2pi

2 cos (t) - 1 = 0, 0 <, = t \leq 2 π

No Solution for t \in R

Solution steps

2 cos (t) - 1 = 0, 0, = t \leq 2 π

Move

1

to the right side

2 cos (t) - 1 = 0

Add

1

to both sides

2 cos (t) - 1 + 1 = 0 + 1

Simplify

2 cos (t) = 1

2 cos (t) = 1

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

0

No Solution for t \in R

2 sin (2 (θ + \frac{π}{6})) = - 2

tan (x) = 0.2092

sin (θ + 1) = cos (θ)

0 = sin (x + c)

7 sin (2 θ) - 2 sin (θ) = 0

What is the general solution for 2cos(t)-1=0,0<,=t<= 2pi ?
The general solution for 2cos(t)-1=0,0<,=t<= 2pi is No Solution for t\in\mathbb{R}