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Popular >

cos(2t)-cos(t)=-0.5,0<t<2pi

Solution

cos (2 t) - cos (t) = - 0.5, 0 < t < 2 π

Solution

t = 0.62831 \dots, t = 2 π - 0.62831 \dots, t = 1.88495 \dots, t = - 1.88495 \dots + 2 π

Degrees

t = 3 6^{\circ}, t = 32 4^{\circ}, t = 10 8^{\circ}, t = 25 2^{\circ}

Solution steps

cos (2 t) - cos (t) = - 0.5, 0 < t < 2 π

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) - 0.5

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

Group like terms

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

Add/Subtract the numbers:

0.5 - 1 = - 0.5

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

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

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

Solve by substitution

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

Let:

cos (t) = u

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

- 0.5 - u + 2 u^{2} = 0 : u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

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

Multiply both sides by

10

- 0.5 - 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

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

Refine

- 5 - 10 u + 20 u^{2} = 0

- 5 - 10 u + 20 u^{2} = 0

Write in the standard form

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

20 u^{2} - 10 u - 5 = 0

Solve with the quadratic formula

20 u^{2} - 10 u - 5 = 0

Quadratic Equation Formula:

For

a = 20, b = - 10, c = - 5

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 20 ( - 5 )}{2 \cdot 20}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 20 ( - 5 )}{2 \cdot 20}

(- 10)^{2} - 4 \cdot 20 (- 5) = 105

(- 10)^{2} - 4 \cdot 20 (- 5)

Apply rule

- (- a) = a

= (- 10)^{2} + 4 \cdot 20 \cdot 5

Apply exponent rule:

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

n

is even

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

= 1 0^{2} + 4 \cdot 20 \cdot 5

Multiply the numbers:

4 \cdot 20 \cdot 5 = 400

= 1 0^{2} + 400

1 0^{2} = 100

= 100 + 400

Add the numbers:

100 + 400 = 500

= 500

Prime factorization of

500 : 2^{2} \cdot 5^{3}

500

500

divides by

2500 = 250 \cdot 2

= 2 \cdot 250

250

divides by

2250 = 125 \cdot 2

= 2 \cdot 2 \cdot 125

125

divides by

5125 = 25 \cdot 5

= 2 \cdot 2 \cdot 5 \cdot 25

25

divides by

525 = 5 \cdot 5

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

2, 5

are all prime numbers, therefore no further factorization is possible

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

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

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

Apply exponent rule:

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

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

Apply radical rule:

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

Apply radical rule:

2^{2} = 2

= 25 5^{2}

Apply radical rule:

5^{2} = 5

= 2 \cdot 55

Refine

= 105

u_{1, 2} = \frac{- ( - 10 ) \pm 10 5}{2 \cdot 20}

Separate the solutions

u_{1} = \frac{- ( - 10 ) + 10 5}{2 \cdot 20}, u_{2} = \frac{- ( - 10 ) - 10 5}{2 \cdot 20}

u = \frac{- ( - 10 ) + 10 5}{2 \cdot 20} : \frac{1 + 5}{4}

\frac{- ( - 10 ) + 10 5}{2 \cdot 20}

Apply rule

- (- a) = a

= \frac{10 + 10 5}{2 \cdot 20}

Multiply the numbers:

2 \cdot 20 = 40

= \frac{10 + 10 5}{40}

Factor

10 + 105 : 10 (1 + 5)

10 + 105

Rewrite as

= 10 \cdot 1 + 105

Factor out common term

10

= 10 (1 + 5)

= \frac{10 ( 1 + 5 )}{40}

Cancel the common factor:

10

= \frac{1 + 5}{4}

u = \frac{- ( - 10 ) - 10 5}{2 \cdot 20} : \frac{1 - 5}{4}

\frac{- ( - 10 ) - 10 5}{2 \cdot 20}

Apply rule

- (- a) = a

= \frac{10 - 10 5}{2 \cdot 20}

Multiply the numbers:

2 \cdot 20 = 40

= \frac{10 - 10 5}{40}

Factor

10 - 105 : 10 (1 - 5)

10 - 105

Rewrite as

= 10 \cdot 1 - 105

Factor out common term

10

= 10 (1 - 5)

= \frac{10 ( 1 - 5 )}{40}

Cancel the common factor:

10

= \frac{1 - 5}{4}

The solutions to the quadratic equation are:

u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

Substitute back

u = cos (t)

cos (t) = \frac{1 + 5}{4}, cos (t) = \frac{1 - 5}{4}

cos (t) = \frac{1 + 5}{4}, cos (t) = \frac{1 - 5}{4}

cos (t) = \frac{1 + 5}{4}, 0 < t < 2 π : t = arccos (\frac{1 + 5}{4}), t = 2 π - arccos (\frac{1 + 5}{4})

cos (t) = \frac{1 + 5}{4}, 0 < t < 2 π

Apply trig inverse properties

cos (t) = \frac{1 + 5}{4}

General solutions for

cos (t) = \frac{1 + 5}{4}

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

t = arccos (\frac{1 + 5}{4}) + 2 πn, t = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

t = arccos (\frac{1 + 5}{4}) + 2 πn, t = 2 π - arccos (\frac{1 + 5}{4}) + 2 πn

Solutions for the range

0 < t < 2 π

t = arccos (\frac{1 + 5}{4}), t = 2 π - arccos (\frac{1 + 5}{4})

cos (t) = \frac{1 - 5}{4}, 0 < t < 2 π : t = arccos (\frac{1 - 5}{4}), t = - arccos (\frac{1 - 5}{4}) + 2 π

cos (t) = \frac{1 - 5}{4}, 0 < t < 2 π

Apply trig inverse properties

cos (t) = \frac{1 - 5}{4}

General solutions for

cos (t) = \frac{1 - 5}{4}

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

t = arccos (\frac{1 - 5}{4}) + 2 πn, t = - arccos (\frac{1 - 5}{4}) + 2 πn

t = arccos (\frac{1 - 5}{4}) + 2 πn, t = - arccos (\frac{1 - 5}{4}) + 2 πn

Solutions for the range

0 < t < 2 π

t = arccos (\frac{1 - 5}{4}), t = - arccos (\frac{1 - 5}{4}) + 2 π

Combine all the solutions

t = arccos (\frac{1 + 5}{4}), t = 2 π - arccos (\frac{1 + 5}{4}), t = arccos (\frac{1 - 5}{4}), t = - arccos (\frac{1 - 5}{4}) + 2 π

Show solutions in decimal form

t = 0.62831 \dots, t = 2 π - 0.62831 \dots, t = 1.88495 \dots, t = - 1.88495 \dots + 2 π

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