What is the general solution for 2sin(x)=4sin(x)+1 ?

The general solution for 2sin(x)=4sin(x)+1 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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2sin(x)=4sin(x)+1

2 sin (x) = 4 sin (x) + 1

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Degrees

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (x) = 4 sin (x) + 1

Solve by substitution

2 sin (x) = 4 sin (x) + 1

Let:

sin (x) = u

2 u = 4 u + 1

2 u = 4 u + 1 : u = - \frac{1}{2}

2 u = 4 u + 1

Move

4 u

to the left side

2 u = 4 u + 1

Subtract

4 u

from both sides

2 u - 4 u = 4 u + 1 - 4 u

Simplify

- 2 u = 1

- 2 u = 1

Divide both sides by

- 2

- 2 u = 1

Divide both sides by

- 2

\frac{- 2 u}{- 2} = \frac{1}{- 2}

Simplify

u = - \frac{1}{2}

u = - \frac{1}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{1}{2}

sin (x) = - \frac{1}{2}

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

sin (x) = - \frac{1}{2}

General solutions for

sin (x) = - \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}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

- 1 - cos (θ) = - 3 cos (θ)

3 \cdot cos (θ) = 2 - sin (θ)

0.06 = 0.1 \cdot cos (4 x - 1.57)

tan (a) = \frac{20}{15}

sin (x) = - \frac{8}{8 13}

What is the general solution for 2sin(x)=4sin(x)+1 ?
The general solution for 2sin(x)=4sin(x)+1 is x=(7pi)/6+2pin,x=(11pi)/6+2pin