What is the general solution for sin(θ/2)-1=0,0<= θ<= 2pi ?

The general solution for sin(θ/2)-1=0,0<= θ<= 2pi is θ=pi

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sin(θ/2)-1=0,0<= θ<= 2pi

sin (\frac{θ}{2}) - 1 = 0, 0 \leq θ \leq 2 π

θ = π

Degrees

θ = 18 0^{\circ}

Solution steps

sin (\frac{θ}{2}) - 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

sin (\frac{θ}{2}) - 1 = 0

Add

1

to both sides

sin (\frac{θ}{2}) - 1 + 1 = 0 + 1

Simplify

sin (\frac{θ}{2}) = 1

sin (\frac{θ}{2}) = 1

General solutions for

sin (\frac{θ}{2}) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

\frac{θ}{2} = \frac{π}{2} + 2 πn

\frac{θ}{2} = \frac{π}{2} + 2 πn

Solve

\frac{θ}{2} = \frac{π}{2} + 2 πn : θ = π + 4 πn

\frac{θ}{2} = \frac{π}{2} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{2} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{π}{2} + 2 \cdot 2 πn : π + 4 πn

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{2}

Cancel the common factor:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

θ = π + 4 πn

θ = π + 4 πn

θ = π + 4 πn

θ = π + 4 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = π

cos (2 x) = - \frac{2}{2}, 0 \leq x \leq 2 π

\frac{3 + sin ^{2} ( θ )}{cos ( θ ) - 2} = 3 cos (θ)

cos (a) = 0

cos (3 x) = \frac{- 3}{2}

2 cos (2 θ) - 10 sin (θ) - 2 = - 6 sin (θ) - 3

What is the general solution for sin(θ/2)-1=0,0<= θ<= 2pi ?
The general solution for sin(θ/2)-1=0,0<= θ<= 2pi is θ=pi