What is the general solution for |sin(x)|=sin(x)+2 ?

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

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Popular Trigonometry >

|sin(x)|=sin(x)+2

Solution

∣ sin (x) ∣ = sin (x) + 2

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

∣ sin (x) ∣ = sin (x) + 2

Solve by substitution

∣ sin (x) ∣ = sin (x) + 2

Let:

sin (x) = u

∣ u ∣ = u + 2

∣ u ∣ = u + 2 : u = - 1

∣ u ∣ = u + 2

Find positive and negative intervals

Find intervals for

∣ u ∣

u \geq 0

u \geq 0, ∣ u ∣ = u

Rewrite

∣ u ∣

for

u \geq 0 : ∣ u ∣ = u

Apply absolute rule: If

u \geq 0

then

∣ u ∣ = u

∣ u ∣ = u

u < 0

u < 0, ∣ u ∣ = - u

Rewrite

∣ u ∣

for

u < 0 : ∣ u ∣ = - u

Apply absolute rule: If

u < 0

then

∣ u ∣ = - u

∣ u ∣ = - u

Identify the intervals:

u < 0, u \geq 0

∣ u ∣ u < 0 - u \geq 0 +

u < 0, u \geq 0

u < 0, u \geq 0

Solve the inequality for each interval

u < 0, u \geq 0

For

u < 0 : u = - 1

For

u < 0

rewrite

∣ u ∣ = u + 2

- u = u + 2

- u = u + 2 : u = - 1

- u = u + 2

Move

u

to the left side

- u = u + 2

Subtract

u

from both sides

- u - u = u + 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

Combine the intervals

u = - 1 and u < 0

Merge Overlapping Intervals

u = - 1 and u < 0

The intersection of two intervals is the set of numbers which are in both intervals

u = - 1

and

u < 0

u = - 1

u = - 1

For

u \geq 0 :

No Solution

For

u \geq 0

rewrite

∣ u ∣ = u + 2

u = u + 2

u = u + 2 :

No Solution

u = u + 2

Subtract

u

from both sides

u - u = u + 2 - u

Simplify

0 = 2

The sides are not equal

No Solution

Combine the intervals

No Solution and u \geq 0

Merge Overlapping Intervals

No Solution and u \geq 0

The intersection of two intervals is the set of numbers which are in both intervals

No Solution

and

u \geq 0

No Solution

No Solution

Combine Solutions:

u = - 1 or No Solution

u = - 1 or No Solution

u = - 1

Substitute back

u = sin (x)

sin (x) = - 1

sin (x) = - 1

sin (x) = - 1 : x = \frac{3 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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

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