What is the general solution for (sin(x)-1)(sin(x)-5)=0 ?

The general solution for (sin(x)-1)(sin(x)-5)=0 is x= pi/2+2pin

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(sin(x)-1)(sin(x)-5)=0

(sin (x) - 1) (sin (x) - 5) = 0

x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

(sin (x) - 1) (sin (x) - 5) = 0

Solving each part separately

sin (x) - 1 = 0 or sin (x) - 5 = 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) - 5 = 0 :

No Solution

sin (x) - 5 = 0

Move

5

to the right side

sin (x) - 5 = 0

Add

5

to both sides

sin (x) - 5 + 5 = 0 + 5

Simplify

sin (x) = 5

sin (x) = 5

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

(4 sin^{2} (x) - 1) (cos^{2} (x) - 1) = 0

3 sin^{2} (x) - 6 sin (x) = - 3, 0 \leq x \leq 2 π

1 + sin (π \cdot t) = \frac{3}{2}

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

csc (x) = \frac{5}{2}

What is the general solution for (sin(x)-1)(sin(x)-5)=0 ?
The general solution for (sin(x)-1)(sin(x)-5)=0 is x= pi/2+2pin