What is the general solution for (sin(x)-1)(sin(x)-4)=0 ?

The general solution for (sin(x)-1)(sin(x)-4)=0 is x= pi/2+2pin

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(sin(x)-1)(sin(x)-4)=0

(sin (x) - 1) (sin (x) - 4) = 0

x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

(sin (x) - 1) (sin (x) - 4) = 0

Solving each part separately

sin (x) - 1 = 0 or sin (x) - 4 = 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) - 4 = 0 :

No Solution

sin (x) - 4 = 0

Move

4

to the right side

sin (x) - 4 = 0

Add

4

to both sides

sin (x) - 4 + 4 = 0 + 4

Simplify

sin (x) = 4

sin (x) = 4

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

cos^{2} (a) = \frac{1}{5}

cos (2 9^{\circ} + θ) = 0.563

cot (θ) = \frac{18}{19}

1 + cos (2 θ) = 3 sin (θ)

2 cos (x) sin (x) - 3 sin (x) = 0

What is the general solution for (sin(x)-1)(sin(x)-4)=0 ?
The general solution for (sin(x)-1)(sin(x)-4)=0 is x= pi/2+2pin