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Popular Trigonometry >

2cos(3x)+cos(2x)+1=0,0<= x<= 2pi

Solution

2 cos (3 x) + cos (2 x) + 1 = 0, 0 \leq x \leq 2 π

Solution

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0.72273 \dots, x = 2 π - 0.72273 \dots, x = π

Degrees

x = 9 0^{\circ}, x = 27 0^{\circ}, x = 41.40962 \dots^{\circ}, x = 318.59037 \dots^{\circ}, x = 18 0^{\circ}

Solution steps

2 cos (3 x) + cos (2 x) + 1 = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

1 + cos (2 x) + 2 cos (3 x)

Use the Double Angle identity:

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

= 1 + 2 cos^{2} (x) - 1 + 2 cos (3 x)

Simplify

1 + 2 cos^{2} (x) - 1 + 2 cos (3 x) : 2 cos^{2} (x) + 2 cos (3 x)

1 + 2 cos^{2} (x) - 1 + 2 cos (3 x)

Group like terms

= 2 cos^{2} (x) + 2 cos (3 x) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (x) + 2 cos (3 x)

= 2 cos^{2} (x) + 2 cos (3 x)

cos (3 x) = 4 cos^{3} (x) - 3 cos (x)

cos (3 x)

Rewrite using trig identities

cos (3 x)

Rewrite as

= cos (2 x + x)

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Simplify

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x) : cos (x) cos (2 x) - 2 sin^{2} (x) cos (x)

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

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

2 sin (x) cos (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (x) sin (x) = sin^{1 + 1} (x)

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

Add the numbers:

1 + 1 = 2

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

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

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

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

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

Expand

(2 cos^{2} (x) - 1) cos (x) - 2 (1 - cos^{2} (x)) cos (x) : 4 cos^{3} (x) - 3 cos (x)

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

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

Expand

cos (x) (2 cos^{2} (x) - 1) : 2 cos^{3} (x) - cos (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = cos (x), b = 2 cos^{2} (x), c = 1

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

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

Simplify

2 cos^{2} (x) cos (x) - 1 \cdot cos (x) : 2 cos^{3} (x) - cos (x)

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

2 cos^{2} (x) cos (x) = 2 cos^{3} (x)

2 cos^{2} (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos^{2} (x) cos (x) = cos^{2 + 1} (x)

= 2 cos^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (x)

1 \cdot cos (x) = cos (x)

1 cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (x)

= 2 cos^{3} (x) - cos (x)

= 2 cos^{3} (x) - cos (x)

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

Expand

- 2 cos (x) (1 - cos^{2} (x)) : - 2 cos (x) + 2 cos^{3} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = - 2 cos (x), b = 1, c = cos^{2} (x)

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

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 cos (x) + 2 cos^{2} (x) cos (x)

Simplify

- 2 \cdot 1 \cdot cos (x) + 2 cos^{2} (x) cos (x) : - 2 cos (x) + 2 cos^{3} (x)

- 2 \cdot 1 cos (x) + 2 cos^{2} (x) cos (x)

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 cos (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos (x)

2 cos^{2} (x) cos (x) = 2 cos^{3} (x)

2 cos^{2} (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos^{2} (x) cos (x) = cos^{2 + 1} (x)

= 2 cos^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= 2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

Simplify

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x) : 4 cos^{3} (x) - 3 cos (x)

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

Group like terms

= 2 cos^{3} (x) + 2 cos^{3} (x) - cos (x) - 2 cos (x)

Add similar elements:

2 cos^{3} (x) + 2 cos^{3} (x) = 4 cos^{3} (x)

= 4 cos^{3} (x) - cos (x) - 2 cos (x)

Add similar elements:

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

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 2 (4 cos^{3} (x) - 3 cos (x)) + 2 cos^{2} (x)

(- 3 cos (x) + 4 cos^{3} (x)) \cdot 2 + 2 cos^{2} (x) = 0

Solve by substitution

(- 3 cos (x) + 4 cos^{3} (x)) \cdot 2 + 2 cos^{2} (x) = 0

Let:

cos (x) = u

(- 3 u + 4 u^{3}) \cdot 2 + 2 u^{2} = 0

(- 3 u + 4 u^{3}) \cdot 2 + 2 u^{2} = 0 : u = 0, u = \frac{3}{4}, u = - 1

(- 3 u + 4 u^{3}) \cdot 2 + 2 u^{2} = 0

Factor

(- 3 u + 4 u^{3}) \cdot 2 + 2 u^{2} : 2 u (4 u - 3) (u + 1)

(- 3 u + 4 u^{3}) \cdot 2 + 2 u^{2}

Factor out common term

2

= 2 (u^{3} \cdot 4 - 3 u + u^{2})

Factor

4 u^{3} + u^{2} - 3 u : u (4 u - 3) (u + 1)

u^{3} \cdot 4 - 3 u + u^{2}

Factor out common term

u : u (4 u^{2} + u - 3)

4 u^{3} + u^{2} - 3 u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{2} = uu

= 4 u^{2} u + uu - 3 u

Factor out common term

u

= u (4 u^{2} + u - 3)

= u (4 u^{2} + u - 3)

Factor

4 u^{2} + u - 3 : (4 u - 3) (u + 1)

4 u^{2} + u - 3

Write in the standard form

a x^{2} + b x + c

= 4 u^{2} + u - 3

Break the expression into groups

4 u^{2} + u - 3

Definition

Factors of

12 : 1, 2, 3, 4, 6, 12

12

Divisors (Factors)

Find the Prime factors of

12 : 2, 2, 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

Multiply the prime factors of

12 : 4, 6

2 \cdot 2 = 4

2 \cdot 3 = 6

4, 6

4, 6

Add the prime factors:

2, 3

Add 1 and the number

12

itself

1, 12

The factors of

12

1, 2, 3, 4, 6, 12

Negative factors of

12 : - 1, - 2, - 3, - 4, - 6, - 12

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 3, - 4, - 6, - 12

For every two factors such that

u * v = - 12,

check if

u + v = 1

Check

u = 1, v = - 12 : u * v = - 12, u + v = - 11 \Rightarrow

FalseCheck

u = 2, v = - 6 : u * v = - 12, u + v = - 4 \Rightarrow

False

u = 4, v = - 3

Group into

(a x^{2} + ux) + (vx + c)

(4 u^{2} - 3 u) + (4 u - 3)

= (4 u^{2} - 3 u) + (4 u - 3)

Factor out

u

from

4 u^{2} - 3 u : u (4 u - 3)

4 u^{2} - 3 u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{2} = uu

= 4 uu - 3 u

Factor out common term

u

= u (4 u - 3)

= u (4 u - 3) + (4 u - 3)

Factor out common term

4 u - 3

= (4 u - 3) (u + 1)

= u (4 u - 3) (u + 1)

= 2 u (u + 1) (4 u - 3)

= 2 u (4 u - 3) (u + 1)

2 u (4 u - 3) (u + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or 4 u - 3 = 0 or u + 1 = 0

Solve

4 u - 3 = 0 : u = \frac{3}{4}

4 u - 3 = 0

Move

3

to the right side

4 u - 3 = 0

Add

3

to both sides

4 u - 3 + 3 = 0 + 3

Simplify

4 u = 3

4 u = 3

Divide both sides by

4

4 u = 3

Divide both sides by

4

\frac{4 u}{4} = \frac{3}{4}

Simplify

u = \frac{3}{4}

u = \frac{3}{4}

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

The solutions are

u = 0, u = \frac{3}{4}, u = - 1

Substitute back

u = cos (x)

cos (x) = 0, cos (x) = \frac{3}{4}, cos (x) = - 1

cos (x) = 0, cos (x) = \frac{3}{4}, cos (x) = - 1

cos (x) = 0, 0 \leq x \leq 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x \leq 2 π

General solutions for

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = \frac{3}{4}, 0 \leq x \leq 2 π : x = arccos (\frac{3}{4}), x = 2 π - arccos (\frac{3}{4})

cos (x) = \frac{3}{4}, 0 \leq x \leq 2 π

Apply trig inverse properties

cos (x) = \frac{3}{4}

General solutions for

cos (x) = \frac{3}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

x = arccos (\frac{3}{4}) + 2 πn, x = 2 π - arccos (\frac{3}{4}) + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = arccos (\frac{3}{4}), x = 2 π - arccos (\frac{3}{4})

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

cos (x) = - 1, 0 \leq x \leq 2 π

General solutions for

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = π

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}, x = arccos (\frac{3}{4}), x = 2 π - arccos (\frac{3}{4}), x = π

Show solutions in decimal form

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0.72273 \dots, x = 2 π - 0.72273 \dots, x = π

Graph

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