English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-4cos(2θ)-3cos(θ)-4=-2cos(θ)-7

Solution

- 4 cos (2 θ) - 3 cos (θ) - 4 = - 2 cos (θ) - 7

Solution

θ = π + 2 πn, θ = 0.50536 \dots + 2 πn, θ = 2 π - 0.50536 \dots + 2 πn

Degrees

θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 28.95502 \dots^{\circ} + 36 0^{\circ} n, θ = 331.04497 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 4 cos (2 θ) - 3 cos (θ) - 4 = - 2 cos (θ) - 7

Subtract

- 2 cos (θ) - 7

from both sides

- 4 cos (2 θ) - cos (θ) + 3 = 0

Rewrite using trig identities

3 - cos (θ) - 4 cos (2 θ)

Use the Double Angle identity:

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

= 3 - cos (θ) - 4 (2 cos^{2} (θ) - 1)

Simplify

3 - cos (θ) - 4 (2 cos^{2} (θ) - 1) : - 8 cos^{2} (θ) - cos (θ) + 7

3 - cos (θ) - 4 (2 cos^{2} (θ) - 1)

Expand

- 4 (2 cos^{2} (θ) - 1) : - 8 cos^{2} (θ) + 4

- 4 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 2 cos^{2} (θ), c = 1

= - 4 \cdot 2 cos^{2} (θ) - (- 4) \cdot 1

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 2 cos^{2} (θ) + 4 \cdot 1

Simplify

- 4 \cdot 2 cos^{2} (θ) + 4 \cdot 1 : - 8 cos^{2} (θ) + 4

- 4 \cdot 2 cos^{2} (θ) + 4 \cdot 1

Multiply the numbers:

4 \cdot 2 = 8

= - 8 cos^{2} (θ) + 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= - 8 cos^{2} (θ) + 4

= - 8 cos^{2} (θ) + 4

= 3 - cos (θ) - 8 cos^{2} (θ) + 4

Simplify

3 - cos (θ) - 8 cos^{2} (θ) + 4 : - 8 cos^{2} (θ) - cos (θ) + 7

3 - cos (θ) - 8 cos^{2} (θ) + 4

Group like terms

= - cos (θ) - 8 cos^{2} (θ) + 3 + 4

Add the numbers:

3 + 4 = 7

= - 8 cos^{2} (θ) - cos (θ) + 7

= - 8 cos^{2} (θ) - cos (θ) + 7

= - 8 cos^{2} (θ) - cos (θ) + 7

7 - cos (θ) - 8 cos^{2} (θ) = 0

Solve by substitution

7 - cos (θ) - 8 cos^{2} (θ) = 0

Let:

cos (θ) = u

7 - u - 8 u^{2} = 0

7 - u - 8 u^{2} = 0 : u = - 1, u = \frac{7}{8}

7 - u - 8 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 8 u^{2} - u + 7 = 0

Solve with the quadratic formula

- 8 u^{2} - u + 7 = 0

Quadratic Equation Formula:

For

a = - 8, b = - 1, c = 7

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 8 ) \cdot 7}{2 ( - 8 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 8 ) \cdot 7}{2 ( - 8 )}

(- 1)^{2} - 4 (- 8) \cdot 7 = 15

(- 1)^{2} - 4 (- 8) \cdot 7

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 8 \cdot 7

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 8 \cdot 7 = 224

4 \cdot 8 \cdot 7

Multiply the numbers:

4 \cdot 8 \cdot 7 = 224

= 224

= 1 + 224

Add the numbers:

1 + 224 = 225

= 225

Factor the number:

225 = 1 5^{2}

= 1 5^{2}

Apply radical rule:

1 5^{2} = 15

= 15

u_{1, 2} = \frac{- ( - 1 ) \pm 15}{2 ( - 8 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 15}{2 ( - 8 )}, u_{2} = \frac{- ( - 1 ) - 15}{2 ( - 8 )}

u = \frac{- ( - 1 ) + 15}{2 ( - 8 )} : - 1

\frac{- ( - 1 ) + 15}{2 ( - 8 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 + 15}{- 2 \cdot 8}

Add the numbers:

1 + 15 = 16

= \frac{16}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{16}{- 16}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{16}{16}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 15}{2 ( - 8 )} : \frac{7}{8}

\frac{- ( - 1 ) - 15}{2 ( - 8 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 - 15}{- 2 \cdot 8}

Subtract the numbers:

1 - 15 = - 14

= \frac{- 14}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 14}{- 16}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{14}{16}

Cancel the common factor:

2

= \frac{7}{8}

The solutions to the quadratic equation are:

u = - 1, u = \frac{7}{8}

Substitute back

u = cos (θ)

cos (θ) = - 1, cos (θ) = \frac{7}{8}

cos (θ) = - 1, cos (θ) = \frac{7}{8}

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

General solutions for

cos (θ) = - 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}

θ = π + 2 πn

θ = π + 2 πn

cos (θ) = \frac{7}{8} : θ = arccos (\frac{7}{8}) + 2 πn, θ = 2 π - arccos (\frac{7}{8}) + 2 πn

cos (θ) = \frac{7}{8}

Apply trig inverse properties

cos (θ) = \frac{7}{8}

General solutions for

cos (θ) = \frac{7}{8}

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

θ = arccos (\frac{7}{8}) + 2 πn, θ = 2 π - arccos (\frac{7}{8}) + 2 πn

θ = arccos (\frac{7}{8}) + 2 πn, θ = 2 π - arccos (\frac{7}{8}) + 2 πn

Combine all the solutions

θ = π + 2 πn, θ = arccos (\frac{7}{8}) + 2 πn, θ = 2 π - arccos (\frac{7}{8}) + 2 πn

Show solutions in decimal form

θ = π + 2 πn, θ = 0.50536 \dots + 2 πn, θ = 2 π - 0.50536 \dots + 2 πn

Popular Examples

sin(x)+sin(2x)=sin(3x)

sin (x) + sin (2 x) = sin (3 x)

2cos^2(x)-9cos(x)+4=0

2 cos^{2} (x) - 9 cos (x) + 4 = 0

solvefor x,sin^2(x)=3cos^2(x)solve for

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

solvefor x,-2sin(2x)=0 solve for

x, - 2 sin (2 x) = 0

sin(θ)+cos(2θ)=1

sin (θ) + cos (2 θ) = 1