What is the general solution for -6sin(θ)+4=3sqrt(3)+4 ?

The general solution for -6sin(θ)+4=3sqrt(3)+4 is θ=(4pi)/3+2pin,θ=(5pi)/3+2pin

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-6sin(θ)+4=3sqrt(3)+4

- 6 sin (θ) + 4 = 33 + 4

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Degrees

θ = 24 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

- 6 sin (θ) + 4 = 33 + 4

Subtract

4

from both sides

- 6 sin (θ) + 4 - 4 = 33 + 4 - 4

Simplify

- 6 sin (θ) = 33

Divide both sides by

- 6

- 6 sin (θ) = 33

Divide both sides by

- 6

\frac{- 6 sin ( θ )}{- 6} = \frac{3 3}{- 6}

Simplify

\frac{- 6 sin ( θ )}{- 6} = \frac{3 3}{- 6}

Simplify

\frac{- 6 sin ( θ )}{- 6} : sin (θ)

\frac{- 6 sin ( θ )}{- 6}

Apply the fraction rule:

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

= \frac{6 sin ( θ )}{6}

Divide the numbers:

\frac{6}{6} = 1

= sin (θ)

Simplify

\frac{3 3}{- 6} : - \frac{3}{2}

\frac{3 3}{- 6}

Apply the fraction rule:

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

= - \frac{3 3}{6}

Cancel the common factor:

3

= - \frac{3}{2}

sin (θ) = - \frac{3}{2}

sin (θ) = - \frac{3}{2}

sin (θ) = - \frac{3}{2}

General solutions for

sin (θ) = - \frac{3}{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{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

- 6 sin (x) + 3 sin (2 x) = 0

- sin (2 θ) - cos (4 θ) = 0, 0 \leq θ \leq 2 π

x, cos^{2} (x) - y^{2} + 3 = 0

cot (x) + 1 = 0, 0 \leq x \leq 2 π

- 4 cos (2 θ) + 7 cos (θ) + 4 = 9 cos (θ) + 7

What is the general solution for -6sin(θ)+4=3sqrt(3)+4 ?
The general solution for -6sin(θ)+4=3sqrt(3)+4 is θ=(4pi)/3+2pin,θ=(5pi)/3+2pin