What is the general solution for 2sin(x)-3sin(x)+1=0 ?

The general solution for 2sin(x)-3sin(x)+1=0 is x= pi/2+2pin

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2sin(x)-3sin(x)+1=0

2 sin (x) - 3 sin (x) + 1 = 0

x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

2 sin (x) - 3 sin (x) + 1 = 0

Simplify

2 sin (x) - 3 sin (x) + 1 : - sin (x) + 1

2 sin (x) - 3 sin (x) + 1

Add similar elements:

2 sin (x) - 3 sin (x) = - sin (x)

= - sin (x) + 1

- sin (x) + 1 = 0

Move

1

to the right side

- sin (x) + 1 = 0

Subtract

1

from both sides

- sin (x) + 1 - 1 = 0 - 1

Simplify

- sin (x) = - 1

- sin (x) = - 1

Divide both sides by

- 1

- sin (x) = - 1

Divide both sides by

- 1

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

Simplify

sin (x) = 1

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

2 sin (x) (sin (x) - 1) = 3 cos^{2} (x)

2 sin^{2} (x) = 2 sin (x) + 4

sin (x) cos^{2} (x) = sin (x), 0 \leq x \leq 2 π

2 \cdot cos (4 x) - 1 = 0

2 cos^{2} (θ) - 3 cos (θ) = 2

What is the general solution for 2sin(x)-3sin(x)+1=0 ?
The general solution for 2sin(x)-3sin(x)+1=0 is x= pi/2+2pin