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Popular >

cos(a)+cos(2a)=-0.75

Solution

cos (a) + cos (2 a) = - 0.75

Solution

a = 1.38674 \dots + 2 πn, a = 2 π - 1.38674 \dots + 2 πn, a = 2.32267 \dots + 2 πn, a = - 2.32267 \dots + 2 πn

Degrees

a = 79.45470 \dots^{\circ} + 36 0^{\circ} n, a = 280.54529 \dots^{\circ} + 36 0^{\circ} n, a = 133.07951 \dots^{\circ} + 36 0^{\circ} n, a = - 133.07951 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos (a) + cos (2 a) = - 0.75

Subtract

- 0.75

from both sides

cos (a) + cos (2 a) + 0.75 = 0

Rewrite using trig identities

0.75 + cos (2 a) + cos (a)

Use the Double Angle identity:

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

= 0.75 + 2 cos^{2} (a) - 1 + cos (a)

Simplify

0.75 + 2 cos^{2} (a) - 1 + cos (a) : 2 cos^{2} (a) + cos (a) - 0.25

0.75 + 2 cos^{2} (a) - 1 + cos (a)

Group like terms

= 2 cos^{2} (a) + cos (a) + 0.75 - 1

Add/Subtract the numbers:

0.75 - 1 = - 0.25

= 2 cos^{2} (a) + cos (a) - 0.25

= 2 cos^{2} (a) + cos (a) - 0.25

- 0.25 + cos (a) + 2 cos^{2} (a) = 0

Solve by substitution

- 0.25 + cos (a) + 2 cos^{2} (a) = 0

Let:

cos (a) = u

- 0.25 + u + 2 u^{2} = 0

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

- 0.25 + u + 2 u^{2} = 0

Multiply both sides by

100

- 0.25 + u + 2 u^{2} = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

- 0.25 \cdot 100 + u \cdot 100 + 2 u^{2} \cdot 100 = 0 \cdot 100

Refine

- 25 + 100 u + 200 u^{2} = 0

- 25 + 100 u + 200 u^{2} = 0

Write in the standard form

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

200 u^{2} + 100 u - 25 = 0

Solve with the quadratic formula

200 u^{2} + 100 u - 25 = 0

Quadratic Equation Formula:

For

a = 200, b = 100, c = - 25

u_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 \cdot 200 ( - 25 )}{2 \cdot 200}

u_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 \cdot 200 ( - 25 )}{2 \cdot 200}

10 0^{2} - 4 \cdot 200 (- 25) = 1003

10 0^{2} - 4 \cdot 200 (- 25)

Apply rule

- (- a) = a

= 10 0^{2} + 4 \cdot 200 \cdot 25

Multiply the numbers:

4 \cdot 200 \cdot 25 = 20000

= 10 0^{2} + 20000

10 0^{2} = 10000

= 10000 + 20000

Add the numbers:

10000 + 20000 = 30000

= 30000

Prime factorization of

30000 : 2^{4} \cdot 3 \cdot 5^{4}

30000

30000

divides by

230000 = 15000 \cdot 2

= 2 \cdot 15000

15000

divides by

215000 = 7500 \cdot 2

= 2 \cdot 2 \cdot 7500

7500

divides by

27500 = 3750 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 3750

3750

divides by

23750 = 1875 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 1875

1875

divides by

31875 = 625 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 625

625

divides by

5625 = 125 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 125

125

divides by

5125 = 25 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 5 \cdot 25

25

divides by

525 = 5 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 5 \cdot 5 \cdot 5

2, 3, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 5 \cdot 5 \cdot 5

= 2^{4} \cdot 3 \cdot 5^{4}

= 2^{4} \cdot 5^{4} \cdot 3

Apply radical rule:

= 3 2^{4} 5^{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 3 5^{4}

Apply radical rule:

5^{4} = 5^{\frac{4}{2}} = 5^{2}

= 2^{2} \cdot 5^{2} 3

Refine

= 1003

u_{1, 2} = \frac{- 100 \pm 100 3}{2 \cdot 200}

Separate the solutions

u_{1} = \frac{- 100 + 100 3}{2 \cdot 200}, u_{2} = \frac{- 100 - 100 3}{2 \cdot 200}

u = \frac{- 100 + 100 3}{2 \cdot 200} : \frac{- 1 + 3}{4}

\frac{- 100 + 100 3}{2 \cdot 200}

Multiply the numbers:

2 \cdot 200 = 400

= \frac{- 100 + 100 3}{400}

Factor

- 100 + 1003 : 100 (- 1 + 3)

- 100 + 1003

Rewrite as

= - 100 \cdot 1 + 1003

Factor out common term

100

= 100 (- 1 + 3)

= \frac{100 ( - 1 + 3 )}{400}

Cancel the common factor:

100

= \frac{- 1 + 3}{4}

u = \frac{- 100 - 100 3}{2 \cdot 200} : - \frac{1 + 3}{4}

\frac{- 100 - 100 3}{2 \cdot 200}

Multiply the numbers:

2 \cdot 200 = 400

= \frac{- 100 - 100 3}{400}

Factor

- 100 - 1003 : - 100 (1 + 3)

- 100 - 1003

Rewrite as

= - 100 \cdot 1 - 1003

Factor out common term

100

= - 100 (1 + 3)

= - \frac{100 ( 1 + 3 )}{400}

Cancel the common factor:

100

= - \frac{1 + 3}{4}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (a)

cos (a) = \frac{- 1 + 3}{4}, cos (a) = - \frac{1 + 3}{4}

cos (a) = \frac{- 1 + 3}{4}, cos (a) = - \frac{1 + 3}{4}

cos (a) = \frac{- 1 + 3}{4} : a = arccos (\frac{- 1 + 3}{4}) + 2 πn, a = 2 π - arccos (\frac{- 1 + 3}{4}) + 2 πn

cos (a) = \frac{- 1 + 3}{4}

Apply trig inverse properties

cos (a) = \frac{- 1 + 3}{4}

General solutions for

cos (a) = \frac{- 1 + 3}{4}

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

a = arccos (\frac{- 1 + 3}{4}) + 2 πn, a = 2 π - arccos (\frac{- 1 + 3}{4}) + 2 πn

a = arccos (\frac{- 1 + 3}{4}) + 2 πn, a = 2 π - arccos (\frac{- 1 + 3}{4}) + 2 πn

cos (a) = - \frac{1 + 3}{4} : a = arccos (- \frac{1 + 3}{4}) + 2 πn, a = - arccos (- \frac{1 + 3}{4}) + 2 πn

cos (a) = - \frac{1 + 3}{4}

Apply trig inverse properties

cos (a) = - \frac{1 + 3}{4}

General solutions for

cos (a) = - \frac{1 + 3}{4}

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

a = arccos (- \frac{1 + 3}{4}) + 2 πn, a = - arccos (- \frac{1 + 3}{4}) + 2 πn

a = arccos (- \frac{1 + 3}{4}) + 2 πn, a = - arccos (- \frac{1 + 3}{4}) + 2 πn

Combine all the solutions

a = arccos (\frac{- 1 + 3}{4}) + 2 πn, a = 2 π - arccos (\frac{- 1 + 3}{4}) + 2 πn, a = arccos (- \frac{1 + 3}{4}) + 2 πn, a = - arccos (- \frac{1 + 3}{4}) + 2 πn

Show solutions in decimal form

a = 1.38674 \dots + 2 πn, a = 2 π - 1.38674 \dots + 2 πn, a = 2.32267 \dots + 2 πn, a = - 2.32267 \dots + 2 πn

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