What is the general solution for 2-4sin(θ)=0 ?

The general solution for 2-4sin(θ)=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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2-4sin(θ)=0

2 - 4 sin (θ) = 0

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

2 - 4 sin (θ) = 0

Move

2

to the right side

2 - 4 sin (θ) = 0

Subtract

2

from both sides

2 - 4 sin (θ) - 2 = 0 - 2

Simplify

- 4 sin (θ) = - 2

- 4 sin (θ) = - 2

Divide both sides by

- 4

- 4 sin (θ) = - 2

Divide both sides by

- 4

\frac{- 4 sin ( θ )}{- 4} = \frac{- 2}{- 4}

Simplify

\frac{- 4 sin ( θ )}{- 4} = \frac{- 2}{- 4}

Simplify

\frac{- 4 sin ( θ )}{- 4} : sin (θ)

\frac{- 4 sin ( θ )}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4 sin ( θ )}{4}

Divide the numbers:

\frac{4}{4} = 1

= sin (θ)

Simplify

\frac{- 2}{- 4} : \frac{1}{2}

\frac{- 2}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

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}

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

2 cos (t) = 1, - π \leq t \leq π

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

What is the general solution for 2-4sin(θ)=0 ?
The general solution for 2-4sin(θ)=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin