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

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

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(sin(x)-1)*(sin(x)-3)=0

(sin (x) - 1) \cdot (sin (x) - 3) = 0

x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

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

Solving each part separately

sin (x) - 1 = 0 or sin (x) - 3 = 0

sin (x) - 1 = 0 : x = \frac{π}{2} + 2 πn

sin (x) - 1 = 0

Move

1

to the right side

sin (x) - 1 = 0

Add

1

to both sides

sin (x) - 1 + 1 = 0 + 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

sin (x) - 3 = 0 :

No Solution

sin (x) - 3 = 0

Move

3

to the right side

sin (x) - 3 = 0

Add

3

to both sides

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

Simplify

sin (x) = 3

sin (x) = 3

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

2 - 3 cos (θ) = 0

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

tan (θ) = - \frac{3}{4}, sec (θ)

\frac{12}{sin ( 12 0 ^{\circ} )} = \frac{5}{sin ( x )}

- 3 tan^{2} (x) = - 1

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