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

The general solution for 4sin(x)-1=2sin(x)+1 is x= pi/2+2pin

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

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

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

4 u - 1 = 2 u + 1

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

4 u - 1 = 2 u + 1

Move

1

to the right side

4 u - 1 = 2 u + 1

Add

1

to both sides

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

Simplify

4 u = 2 u + 2

4 u = 2 u + 2

Move

2 u

to the left side

4 u = 2 u + 2

Subtract

2 u

from both sides

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

Simplify

2 u = 2

2 u = 2

Divide both sides by

2

2 u = 2

Divide both sides by

2

\frac{2 u}{2} = \frac{2}{2}

Simplify

u = 1

u = 1

Substitute back

u = sin (x)

sin (x) = 1

sin (x) = 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn

sin (3 x) = \frac{3}{4}

sin (x + \frac{π}{3}) + cos (x + \frac{π}{6}) = \frac{3}{3}

4 sin^{2} (x) cot (x) - 3 cot (x) = 0

sin (2 x) = - 2 cos^{2} (x)

cos (\frac{2 x + 3}{2}) = - \frac{1}{2}

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