What is the general solution for cos(t)-cos(2t)=-0.75 ?

The general solution for cos(t)-cos(2t)=-0.75 is t=2.37207…+2pin,t=-2.37207…+2pin

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Popular >

cos(t)-cos(2t)=-0.75

Solution

cos (t) - cos (2 t) = - 0.75

Solution

t = 2.37207 \dots + 2 πn, t = - 2.37207 \dots + 2 πn

Degrees

t = 135.90984 \dots^{\circ} + 36 0^{\circ} n, t = - 135.90984 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos (t) - cos (2 t) = - 0.75

Subtract

- 0.75

from both sides

cos (t) - cos (2 t) + 0.75 = 0

Rewrite using trig identities

0.75 - cos (2 t) + cos (t)

Use the Double Angle identity:

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

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

Simplify

0.75 - (2 cos^{2} (t) - 1) + cos (t) : cos (t) - 2 cos^{2} (t) + 1.75

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

Simplify

0.75 - 2 cos^{2} (t) + 1 + cos (t) : cos (t) - 2 cos^{2} (t) + 1.75

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

Group like terms

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

Add the numbers:

0.75 + 1 = 1.75

= cos (t) - 2 cos^{2} (t) + 1.75

= cos (t) - 2 cos^{2} (t) + 1.75

= cos (t) - 2 cos^{2} (t) + 1.75

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

Solve by substitution

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

Let:

cos (t) = u

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

1.75 + u - 2 u^{2} = 0 : u = - \frac{- 1 + 15}{4}, u = \frac{1 + 15}{4}

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

Multiply both sides by

100

1.75 + 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

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

Refine

175 + 100 u - 200 u^{2} = 0

175 + 100 u - 200 u^{2} = 0

Write in the standard form

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

- 200 u^{2} + 100 u + 175 = 0

Solve with the quadratic formula

- 200 u^{2} + 100 u + 175 = 0

Quadratic Equation Formula:

For

a = - 200, b = 100, c = 175

u_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 ( - 200 ) \cdot 175}{2 ( - 200 )}

u_{1, 2} = \frac{- 100 \pm 10 0 ^{2} - 4 ( - 200 ) \cdot 175}{2 ( - 200 )}

10 0^{2} - 4 (- 200) \cdot 175 = 10015

10 0^{2} - 4 (- 200) \cdot 175

Apply rule

- (- a) = a

= 10 0^{2} + 4 \cdot 200 \cdot 175

Multiply the numbers:

4 \cdot 200 \cdot 175 = 140000

= 10 0^{2} + 140000

10 0^{2} = 10000

= 10000 + 140000

Add the numbers:

10000 + 140000 = 150000

= 150000

Prime factorization of

150000 : 2^{4} \cdot 3 \cdot 5^{5}

150000

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

Apply exponent rule:

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

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

Apply radical rule:

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

Apply radical rule:

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

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

Apply radical rule:

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

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

Refine

= 10015

u_{1, 2} = \frac{- 100 \pm 100 15}{2 ( - 200 )}

Separate the solutions

u_{1} = \frac{- 100 + 100 15}{2 ( - 200 )}, u_{2} = \frac{- 100 - 100 15}{2 ( - 200 )}

u = \frac{- 100 + 100 15}{2 ( - 200 )} : - \frac{- 1 + 15}{4}

\frac{- 100 + 100 15}{2 ( - 200 )}

Remove parentheses:

(- a) = - a

= \frac{- 100 + 100 15}{- 2 \cdot 200}

Multiply the numbers:

2 \cdot 200 = 400

= \frac{- 100 + 100 15}{- 400}

Apply the fraction rule:

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

= - \frac{- 100 + 100 15}{400}

Cancel

\frac{- 100 + 100 15}{400} : \frac{15 - 1}{4}

\frac{- 100 + 100 15}{400}

Factor

- 100 + 10015 : 100 (- 1 + 15)

- 100 + 10015

Rewrite as

= - 100 \cdot 1 + 10015

Factor out common term

100

= 100 (- 1 + 15)

= \frac{100 ( - 1 + 15 )}{400}

Cancel the common factor:

100

= \frac{- 1 + 15}{4}

= - \frac{15 - 1}{4}

= - \frac{- 1 + 15}{4}

u = \frac{- 100 - 100 15}{2 ( - 200 )} : \frac{1 + 15}{4}

\frac{- 100 - 100 15}{2 ( - 200 )}

Remove parentheses:

(- a) = - a

= \frac{- 100 - 100 15}{- 2 \cdot 200}

Multiply the numbers:

2 \cdot 200 = 400

= \frac{- 100 - 100 15}{- 400}

Apply the fraction rule:

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

- 100 - 10015 = - (100 + 10015)

= \frac{100 + 100 15}{400}

Factor

100 + 10015 : 100 (1 + 15)

100 + 10015

Rewrite as

= 100 \cdot 1 + 10015

Factor out common term

100

= 100 (1 + 15)

= \frac{100 ( 1 + 15 )}{400}

Cancel the common factor:

100

= \frac{1 + 15}{4}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 15}{4}, u = \frac{1 + 15}{4}

Substitute back

u = cos (t)

cos (t) = - \frac{- 1 + 15}{4}, cos (t) = \frac{1 + 15}{4}

cos (t) = - \frac{- 1 + 15}{4}, cos (t) = \frac{1 + 15}{4}

cos (t) = - \frac{- 1 + 15}{4} : t = arccos (- \frac{- 1 + 15}{4}) + 2 πn, t = - arccos (- \frac{- 1 + 15}{4}) + 2 πn

cos (t) = - \frac{- 1 + 15}{4}

Apply trig inverse properties

cos (t) = - \frac{- 1 + 15}{4}

General solutions for

cos (t) = - \frac{- 1 + 15}{4}

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

t = arccos (- \frac{- 1 + 15}{4}) + 2 πn, t = - arccos (- \frac{- 1 + 15}{4}) + 2 πn

t = arccos (- \frac{- 1 + 15}{4}) + 2 πn, t = - arccos (- \frac{- 1 + 15}{4}) + 2 πn

cos (t) = \frac{1 + 15}{4} :

No Solution

cos (t) = \frac{1 + 15}{4}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

t = arccos (- \frac{- 1 + 15}{4}) + 2 πn, t = - arccos (- \frac{- 1 + 15}{4}) + 2 πn

Show solutions in decimal form

t = 2.37207 \dots + 2 πn, t = - 2.37207 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(t)-cos(2t)=-0.75 ?
The general solution for cos(t)-cos(2t)=-0.75 is t=2.37207…+2pin,t=-2.37207…+2pin