What is the general solution for 3cos(2x)+4cos(4x)-5=0 ?

The general solution for 3cos(2x)+4cos(4x)-5=0 is x=(0.47431…)/2+pin,x=pi-(0.47431…)/2+pin

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Popular >

3cos(2x)+4cos(4x)-5=0

Solution

3 cos (2 x) + 4 cos (4 x) - 5 = 0

Solution

x = \frac{0.47431 \dots}{2} + πn, x = π - \frac{0.47431 \dots}{2} + πn

Degrees

x = 13.58814 \dots^{\circ} + 18 0^{\circ} n, x = 166.41185 \dots^{\circ} + 18 0^{\circ} n

Solution steps

3 cos (2 x) + 4 cos (4 x) - 5 = 0

Rewrite using trig identities

- 5 + 3 cos (2 x) + 4 cos (4 x)

cos (4 x) = 2 cos^{2} (2 x) - 1

cos (4 x)

Rewrite as

= cos (2 \cdot 2 x)

Use the Double Angle identity:

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

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

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

= - 5 + 3 cos (2 x) + 4 (2 cos^{2} (2 x) - 1)

Simplify

- 5 + 3 cos (2 x) + 4 (2 cos^{2} (2 x) - 1) : 8 cos^{2} (2 x) + 3 cos (2 x) - 9

- 5 + 3 cos (2 x) + 4 (2 cos^{2} (2 x) - 1)

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (2 x), c = 1

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

Simplify

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

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

Multiply the numbers:

4 \cdot 2 = 8

= 8 cos^{2} (2 x) - 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= 8 cos^{2} (2 x) - 4

= 8 cos^{2} (2 x) - 4

= - 5 + 3 cos (2 x) + 8 cos^{2} (2 x) - 4

Simplify

- 5 + 3 cos (2 x) + 8 cos^{2} (2 x) - 4 : 8 cos^{2} (2 x) + 3 cos (2 x) - 9

- 5 + 3 cos (2 x) + 8 cos^{2} (2 x) - 4

Group like terms

= 3 cos (2 x) + 8 cos^{2} (2 x) - 5 - 4

Subtract the numbers:

- 5 - 4 = - 9

= 8 cos^{2} (2 x) + 3 cos (2 x) - 9

= 8 cos^{2} (2 x) + 3 cos (2 x) - 9

= 8 cos^{2} (2 x) + 3 cos (2 x) - 9

- 9 + 3 cos (2 x) + 8 cos^{2} (2 x) = 0

Solve by substitution

- 9 + 3 cos (2 x) + 8 cos^{2} (2 x) = 0

Let:

cos (2 x) = u

- 9 + 3 u + 8 u^{2} = 0

- 9 + 3 u + 8 u^{2} = 0 : u = \frac{- 3 + 3 33}{16}, u = \frac{- 3 - 3 33}{16}

- 9 + 3 u + 8 u^{2} = 0

Write in the standard form

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

8 u^{2} + 3 u - 9 = 0

Solve with the quadratic formula

8 u^{2} + 3 u - 9 = 0

Quadratic Equation Formula:

For

a = 8, b = 3, c = - 9

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 8 ( - 9 )}{2 \cdot 8}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 8 ( - 9 )}{2 \cdot 8}

3^{2} - 4 \cdot 8 (- 9) = 333

3^{2} - 4 \cdot 8 (- 9)

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 8 \cdot 9

Multiply the numbers:

4 \cdot 8 \cdot 9 = 288

= 3^{2} + 288

3^{2} = 9

= 9 + 288

Add the numbers:

9 + 288 = 297

= 297

Prime factorization of

297 : 3^{3} \cdot 11

297

297

divides by

3297 = 99 \cdot 3

= 3 \cdot 99

99

divides by

399 = 33 \cdot 3

= 3 \cdot 3 \cdot 33

33

divides by

333 = 11 \cdot 3

= 3 \cdot 3 \cdot 3 \cdot 11

3, 11

are all prime numbers, therefore no further factorization is possible

= 3 \cdot 3 \cdot 3 \cdot 11

= 3^{3} \cdot 11

= 3^{3} \cdot 11

Apply exponent rule:

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

= 3^{2} \cdot 3 \cdot 11

Apply radical rule:

= 3^{2} 3 \cdot 11

Apply radical rule:

3^{2} = 3

= 3 3 \cdot 11

Refine

= 333

u_{1, 2} = \frac{- 3 \pm 3 33}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- 3 + 3 33}{2 \cdot 8}, u_{2} = \frac{- 3 - 3 33}{2 \cdot 8}

u = \frac{- 3 + 3 33}{2 \cdot 8} : \frac{- 3 + 3 33}{16}

\frac{- 3 + 3 33}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 3 + 3 33}{16}

u = \frac{- 3 - 3 33}{2 \cdot 8} : \frac{- 3 - 3 33}{16}

\frac{- 3 - 3 33}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 3 - 3 33}{16}

The solutions to the quadratic equation are:

u = \frac{- 3 + 3 33}{16}, u = \frac{- 3 - 3 33}{16}

Substitute back

u = cos (2 x)

cos (2 x) = \frac{- 3 + 3 33}{16}, cos (2 x) = \frac{- 3 - 3 33}{16}

cos (2 x) = \frac{- 3 + 3 33}{16}, cos (2 x) = \frac{- 3 - 3 33}{16}

cos (2 x) = \frac{- 3 + 3 33}{16} : x = \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn, x = π - \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn

cos (2 x) = \frac{- 3 + 3 33}{16}

Apply trig inverse properties

cos (2 x) = \frac{- 3 + 3 33}{16}

General solutions for

cos (2 x) = \frac{- 3 + 3 33}{16}

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

2 x = arccos (\frac{- 3 + 3 33}{16}) + 2 πn, 2 x = 2 π - arccos (\frac{- 3 + 3 33}{16}) + 2 πn

2 x = arccos (\frac{- 3 + 3 33}{16}) + 2 πn, 2 x = 2 π - arccos (\frac{- 3 + 3 33}{16}) + 2 πn

Solve

2 x = arccos (\frac{- 3 + 3 33}{16}) + 2 πn : x = \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn

2 x = arccos (\frac{- 3 + 3 33}{16}) + 2 πn

Divide both sides by

2

2 x = arccos (\frac{- 3 + 3 33}{16}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn

x = \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn

Solve

2 x = 2 π - arccos (\frac{- 3 + 3 33}{16}) + 2 πn : x = π - \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn

2 x = 2 π - arccos (\frac{- 3 + 3 33}{16}) + 2 πn

Divide both sides by

2

2 x = 2 π - arccos (\frac{- 3 + 3 33}{16}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + \frac{2 πn}{2}

Simplify

x = π - \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn

x = π - \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn

x = \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn, x = π - \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn

cos (2 x) = \frac{- 3 - 3 33}{16} :

No Solution

cos (2 x) = \frac{- 3 - 3 33}{16}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn, x = π - \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn

Show solutions in decimal form

x = \frac{0.47431 \dots}{2} + πn, x = π - \frac{0.47431 \dots}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos(2x)+4cos(4x)-5=0 ?
The general solution for 3cos(2x)+4cos(4x)-5=0 is x=(0.47431…)/2+pin,x=pi-(0.47431…)/2+pin