What is the general solution for solvefor t,x=cos(t)+cos(2t) ?

The general solution for solvefor t,x=cos(t)+cos(2t) is

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Popular >

solvefor t,x=cos(t)+cos(2t)

Solution

solve for

t, x = cos (t) + cos (2 t)

Solution

t = arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn

Solution steps

x = cos (t) + cos (2 t)

Switch sides

cos (t) + cos (2 t) = x

Subtract

x

from both sides

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

Rewrite using trig identities

cos (2 t) + cos (t) - x

Use the Double Angle identity:

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

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

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

Solve by substitution

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

Let:

cos (t) = u

- 1 + u - x + 2 u^{2} = 0

- 1 + u - x + 2 u^{2} = 0 : u = \frac{- 1 + 8 x + 9}{4}, u = \frac{- 1 - 8 x + 9}{4}

- 1 + u - x + 2 u^{2} = 0

Write in the standard form

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

2 u^{2} + u - 1 - x = 0

Solve with the quadratic formula

2 u^{2} + u - 1 - x = 0

Quadratic Equation Formula:

For

a = 2, b = 1, c = - 1 - x

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

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

Simplify

1^{2} - 4 \cdot 2 (- 1 - x) : 8 x + 9

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- x - 1)

Multiply the numbers:

4 \cdot 2 = 8

= 1 - 8 (- x - 1)

Expand

1 - 8 (- 1 - x) : 8 x + 9

1 - 8 (- 1 - x)

Expand

- 8 (- 1 - x) : 8 + 8 x

- 8 (- 1 - x)

Apply the distributive law:

a (b - c) = ab - a c

a = - 8, b = - 1, c = x

= - 8 (- 1) - (- 8) x

Apply minus-plus rules

- (- a) = a

= 8 \cdot 1 + 8 x

Multiply the numbers:

8 \cdot 1 = 8

= 8 + 8 x

= 1 + 8 + 8 x

Add the numbers:

1 + 8 = 9

= 8 x + 9

= 8 x + 9

u_{1, 2} = \frac{- 1 \pm 8 x + 9}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 1 + 8 x + 9}{2 \cdot 2}, u_{2} = \frac{- 1 - 8 x + 9}{2 \cdot 2}

u = \frac{- 1 + 8 x + 9}{2 \cdot 2} : \frac{- 1 + 8 x + 9}{4}

\frac{- 1 + 8 x + 9}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 + 8 x + 9}{4}

u = \frac{- 1 - 8 x + 9}{2 \cdot 2} : \frac{- 1 - 8 x + 9}{4}

\frac{- 1 - 8 x + 9}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 - 8 x + 9}{4}

The solutions to the quadratic equation are:

u = \frac{- 1 + 8 x + 9}{4}, u = \frac{- 1 - 8 x + 9}{4}

Substitute back

u = cos (t)

cos (t) = \frac{- 1 + 8 x + 9}{4}, cos (t) = \frac{- 1 - 8 x + 9}{4}

cos (t) = \frac{- 1 + 8 x + 9}{4}, cos (t) = \frac{- 1 - 8 x + 9}{4}

cos (t) = \frac{- 1 + 8 x + 9}{4} : t = arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn

cos (t) = \frac{- 1 + 8 x + 9}{4}

Apply trig inverse properties

cos (t) = \frac{- 1 + 8 x + 9}{4}

General solutions for

cos (t) = \frac{- 1 + 8 x + 9}{4}

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

t = arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn

t = arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn

cos (t) = \frac{- 1 - 8 x + 9}{4} : t = arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn

cos (t) = \frac{- 1 - 8 x + 9}{4}

Apply trig inverse properties

cos (t) = \frac{- 1 - 8 x + 9}{4}

General solutions for

cos (t) = \frac{- 1 - 8 x + 9}{4}

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

t = arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn

t = arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn

Combine all the solutions

t = arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn

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9 - 9 sin (θ) = 6 cos^{2} (θ)

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sec (2 x) = - 2

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor t,x=cos(t)+cos(2t) ?
The general solution for solvefor t,x=cos(t)+cos(2t) is