What is the general solution for cos(2t)-cos(t)=0.5 ?

The general solution for cos(2t)-cos(t)=0.5 is t=2.28020…+2pin,t=-2.28020…+2pin

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Popular Trigonometry >

cos(2t)-cos(t)=0.5

Solution

cos (2 t) - cos (t) = 0.5

Solution

t = 2.28020 \dots + 2 πn, t = - 2.28020 \dots + 2 πn

Degrees

t = 130.64631 \dots^{\circ} + 36 0^{\circ} n, t = - 130.64631 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos (2 t) - cos (t) = 0.5

Subtract

0.5

from both sides

cos (2 t) - cos (t) - 0.5 = 0

Rewrite using trig identities

- 0.5 + cos (2 t) - cos (t)

Use the Double Angle identity:

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

= - 0.5 + 2 cos^{2} (t) - 1 - cos (t)

Simplify

- 0.5 + 2 cos^{2} (t) - 1 - cos (t) : 2 cos^{2} (t) - cos (t) - 1.5

- 0.5 + 2 cos^{2} (t) - 1 - cos (t)

Group like terms

= 2 cos^{2} (t) - cos (t) - 0.5 - 1

Subtract the numbers:

- 0.5 - 1 = - 1.5

= 2 cos^{2} (t) - cos (t) - 1.5

= 2 cos^{2} (t) - cos (t) - 1.5

- 1.5 - cos (t) + 2 cos^{2} (t) = 0

Solve by substitution

- 1.5 - cos (t) + 2 cos^{2} (t) = 0

Let:

cos (t) = u

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

- 1.5 - u + 2 u^{2} = 0 : u = \frac{1 + 13}{4}, u = \frac{1 - 13}{4}

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

Multiply both sides by

10

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

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

- 1.5 \cdot 10 - u \cdot 10 + 2 u^{2} \cdot 10 = 0 \cdot 10

Refine

- 15 - 10 u + 20 u^{2} = 0

- 15 - 10 u + 20 u^{2} = 0

Write in the standard form

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

20 u^{2} - 10 u - 15 = 0

Solve with the quadratic formula

20 u^{2} - 10 u - 15 = 0

Quadratic Equation Formula:

For

a = 20, b = - 10, c = - 15

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 20 ( - 15 )}{2 \cdot 20}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 20 ( - 15 )}{2 \cdot 20}

(- 10)^{2} - 4 \cdot 20 (- 15) = 1013

(- 10)^{2} - 4 \cdot 20 (- 15)

Apply rule

- (- a) = a

= (- 10)^{2} + 4 \cdot 20 \cdot 15

Apply exponent rule:

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

n

is even

(- 10)^{2} = 1 0^{2}

= 1 0^{2} + 4 \cdot 20 \cdot 15

Multiply the numbers:

4 \cdot 20 \cdot 15 = 1200

= 1 0^{2} + 1200

1 0^{2} = 100

= 100 + 1200

Add the numbers:

100 + 1200 = 1300

= 1300

Prime factorization of

1300 : 2^{2} \cdot 5^{2} \cdot 13

1300

1300

divides by

21300 = 650 \cdot 2

= 2 \cdot 650

650

divides by

2650 = 325 \cdot 2

= 2 \cdot 2 \cdot 325

325

divides by

5325 = 65 \cdot 5

= 2 \cdot 2 \cdot 5 \cdot 65

65

divides by

565 = 13 \cdot 5

= 2 \cdot 2 \cdot 5 \cdot 5 \cdot 13

2, 5, 13

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 5 \cdot 5 \cdot 13

= 2^{2} \cdot 5^{2} \cdot 13

= 2^{2} \cdot 5^{2} \cdot 13

Apply radical rule:

= 13 2^{2} 5^{2}

Apply radical rule:

2^{2} = 2

= 213 5^{2}

Apply radical rule:

5^{2} = 5

= 2 \cdot 513

Refine

= 1013

u_{1, 2} = \frac{- ( - 10 ) \pm 10 13}{2 \cdot 20}

Separate the solutions

u_{1} = \frac{- ( - 10 ) + 10 13}{2 \cdot 20}, u_{2} = \frac{- ( - 10 ) - 10 13}{2 \cdot 20}

u = \frac{- ( - 10 ) + 10 13}{2 \cdot 20} : \frac{1 + 13}{4}

\frac{- ( - 10 ) + 10 13}{2 \cdot 20}

Apply rule

- (- a) = a

= \frac{10 + 10 13}{2 \cdot 20}

Multiply the numbers:

2 \cdot 20 = 40

= \frac{10 + 10 13}{40}

Factor

10 + 1013 : 10 (1 + 13)

10 + 1013

Rewrite as

= 10 \cdot 1 + 1013

Factor out common term

10

= 10 (1 + 13)

= \frac{10 ( 1 + 13 )}{40}

Cancel the common factor:

10

= \frac{1 + 13}{4}

u = \frac{- ( - 10 ) - 10 13}{2 \cdot 20} : \frac{1 - 13}{4}

\frac{- ( - 10 ) - 10 13}{2 \cdot 20}

Apply rule

- (- a) = a

= \frac{10 - 10 13}{2 \cdot 20}

Multiply the numbers:

2 \cdot 20 = 40

= \frac{10 - 10 13}{40}

Factor

10 - 1013 : 10 (1 - 13)

10 - 1013

Rewrite as

= 10 \cdot 1 - 1013

Factor out common term

10

= 10 (1 - 13)

= \frac{10 ( 1 - 13 )}{40}

Cancel the common factor:

10

= \frac{1 - 13}{4}

The solutions to the quadratic equation are:

u = \frac{1 + 13}{4}, u = \frac{1 - 13}{4}

Substitute back

u = cos (t)

cos (t) = \frac{1 + 13}{4}, cos (t) = \frac{1 - 13}{4}

cos (t) = \frac{1 + 13}{4}, cos (t) = \frac{1 - 13}{4}

cos (t) = \frac{1 + 13}{4} :

No Solution

cos (t) = \frac{1 + 13}{4}

- 1 \leq cos (x) \leq 1

No Solution

cos (t) = \frac{1 - 13}{4} : t = arccos (\frac{1 - 13}{4}) + 2 πn, t = - arccos (\frac{1 - 13}{4}) + 2 πn

cos (t) = \frac{1 - 13}{4}

Apply trig inverse properties

cos (t) = \frac{1 - 13}{4}

General solutions for

cos (t) = \frac{1 - 13}{4}

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

t = arccos (\frac{1 - 13}{4}) + 2 πn, t = - arccos (\frac{1 - 13}{4}) + 2 πn

t = arccos (\frac{1 - 13}{4}) + 2 πn, t = - arccos (\frac{1 - 13}{4}) + 2 πn

Combine all the solutions

t = arccos (\frac{1 - 13}{4}) + 2 πn, t = - arccos (\frac{1 - 13}{4}) + 2 πn

Show solutions in decimal form

t = 2.28020 \dots + 2 πn, t = - 2.28020 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2t)-cos(t)=0.5 ?
The general solution for cos(2t)-cos(t)=0.5 is t=2.28020…+2pin,t=-2.28020…+2pin