What is the general solution for cos(x)+|cos(x)|=1,-2pi<= x<= 2pi ?

Q: What is the general solution for cos(x)+|cos(x)|=1,-2pi<= x<= 2pi ?

The general solution for cos(x)+|cos(x)|=1,-2pi<= x<= 2pi is x= pi/3 ,x=(5pi)/3 ,x=-(5pi)/3 ,x=-pi/3

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Popular Trigonometry >

cos(x)+|cos(x)|=1,-2pi<= x<= 2pi

Solution

cos (x) + ∣ cos (x) ∣ = 1, - 2 π \leq x \leq 2 π

Solution

x = \frac{π}{3}, x = \frac{5 π}{3}, x = - \frac{5 π}{3}, x = - \frac{π}{3}

Degrees

x = 6 0^{\circ}, x = 30 0^{\circ}, x = - 30 0^{\circ}, x = - 6 0^{\circ}

Solution steps

cos (x) + ∣ cos (x) ∣ = 1, - 2 π \leq x \leq 2 π

Solve by substitution

cos (x) + ∣ cos (x) ∣ = 1

Let:

cos (x) = u

u + ∣ u ∣ = 1

u + ∣ u ∣ = 1 : u = \frac{1}{2}

u + ∣ u ∣ = 1

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 :

No Solution

For

u < 0

rewrite

u + ∣ u ∣ = 1

u - u = 1

u - u = 1 :

No Solution

u - u = 1

Add similar elements:

u - u = 0

0 = 1

The sides are not equal

No Solution

Combine the intervals

No Solution and u < 0

Merge Overlapping Intervals

No Solution and u < 0

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

No Solution

and

u < 0

No Solution

No Solution

For

u \geq 0 : u = \frac{1}{2}

For

u \geq 0

rewrite

u + ∣ u ∣ = 1

u + u = 1

u + u = 1 : u = \frac{1}{2}

u + u = 1

Add similar elements:

u + u = 2 u

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

Combine the intervals

u = \frac{1}{2} and u \geq 0

Merge Overlapping Intervals

u = \frac{1}{2} and u \geq 0

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

u = \frac{1}{2}

and

u \geq 0

u = \frac{1}{2}

u = \frac{1}{2}

Combine Solutions:

No Solution or u = \frac{1}{2}

No Solution or u = \frac{1}{2}

u = \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}, - 2 π \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}, x = - \frac{5 π}{3}, x = - \frac{π}{3}

cos (x) = \frac{1}{2}, - 2 π \leq x \leq 2 π

General solutions for

cos (x) = \frac{1}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solutions for the range

- 2 π \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}, x = - \frac{5 π}{3}, x = - \frac{π}{3}

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}, x = - \frac{5 π}{3}, x = - \frac{π}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos(x)+|cos(x)|=1,-2pi<= x<= 2pi ?
The general solution for cos(x)+|cos(x)|=1,-2pi<= x<= 2pi is x= pi/3 ,x=(5pi)/3 ,x=-(5pi)/3 ,x=-pi/3