What is the general solution for 1-cos^2(θ)-6959cos(θ)=0 ?

The general solution for 1-cos^2(θ)-6959cos(θ)=0 is θ=1.57065…+2pin,θ=2pi-1.57065…+2pin

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Popular >

1-cos^2(θ)-6959cos(θ)=0

Solution

1 - cos^{2} (θ) - 6959 cos (θ) = 0

Solution

θ = 1.57065 \dots + 2 πn, θ = 2 π - 1.57065 \dots + 2 πn

Degrees

θ = 89.99176 \dots^{\circ} + 36 0^{\circ} n, θ = 270.00823 \dots^{\circ} + 36 0^{\circ} n

Solution steps

1 - cos^{2} (θ) - 6959 cos (θ) = 0

Solve by substitution

1 - cos^{2} (θ) - 6959 cos (θ) = 0

Let:

cos (θ) = u

1 - u^{2} - 6959 u = 0

1 - u^{2} - 6959 u = 0 : u = - \frac{6959 + 48427685}{2}, u = \frac{48427685 - 6959}{2}

1 - u^{2} - 6959 u = 0

Write in the standard form

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

- u^{2} - 6959 u + 1 = 0

Solve with the quadratic formula

- u^{2} - 6959 u + 1 = 0

Quadratic Equation Formula:

For

a = - 1, b = - 6959, c = 1

u_{1, 2} = \frac{- ( - 6959 ) \pm ( - 6959 ) ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 6959 ) \pm ( - 6959 ) ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

(- 6959)^{2} - 4 (- 1) \cdot 1 = 48427685

(- 6959)^{2} - 4 (- 1) \cdot 1

Apply rule

- (- a) = a

= (- 6959)^{2} + 4 \cdot 1 \cdot 1

Apply exponent rule:

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

n

is even

(- 6959)^{2} = 695 9^{2}

= 695 9^{2} + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 695 9^{2} + 4

695 9^{2} = 48427681

= 48427681 + 4

Add the numbers:

48427681 + 4 = 48427685

= 48427685

u_{1, 2} = \frac{- ( - 6959 ) \pm 48427685}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- ( - 6959 ) + 48427685}{2 ( - 1 )}, u_{2} = \frac{- ( - 6959 ) - 48427685}{2 ( - 1 )}

u = \frac{- ( - 6959 ) + 48427685}{2 ( - 1 )} : - \frac{6959 + 48427685}{2}

\frac{- ( - 6959 ) + 48427685}{2 ( - 1 )}

Remove parentheses:

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

= \frac{6959 + 48427685}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6959 + 48427685}{- 2}

Apply the fraction rule:

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

= - \frac{6959 + 48427685}{2}

u = \frac{- ( - 6959 ) - 48427685}{2 ( - 1 )} : \frac{48427685 - 6959}{2}

\frac{- ( - 6959 ) - 48427685}{2 ( - 1 )}

Remove parentheses:

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

= \frac{6959 - 48427685}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6959 - 48427685}{- 2}

Apply the fraction rule:

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

6959 - 48427685 = - (48427685 - 6959)

= \frac{48427685 - 6959}{2}

The solutions to the quadratic equation are:

u = - \frac{6959 + 48427685}{2}, u = \frac{48427685 - 6959}{2}

Substitute back

u = cos (θ)

cos (θ) = - \frac{6959 + 48427685}{2}, cos (θ) = \frac{48427685 - 6959}{2}

cos (θ) = - \frac{6959 + 48427685}{2}, cos (θ) = \frac{48427685 - 6959}{2}

cos (θ) = - \frac{6959 + 48427685}{2} :

No Solution

cos (θ) = - \frac{6959 + 48427685}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = \frac{48427685 - 6959}{2} : θ = arccos (\frac{48427685 - 6959}{2}) + 2 πn, θ = 2 π - arccos (\frac{48427685 - 6959}{2}) + 2 πn

cos (θ) = \frac{48427685 - 6959}{2}

Apply trig inverse properties

cos (θ) = \frac{48427685 - 6959}{2}

General solutions for

cos (θ) = \frac{48427685 - 6959}{2}

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

θ = arccos (\frac{48427685 - 6959}{2}) + 2 πn, θ = 2 π - arccos (\frac{48427685 - 6959}{2}) + 2 πn

θ = arccos (\frac{48427685 - 6959}{2}) + 2 πn, θ = 2 π - arccos (\frac{48427685 - 6959}{2}) + 2 πn

Combine all the solutions

θ = arccos (\frac{48427685 - 6959}{2}) + 2 πn, θ = 2 π - arccos (\frac{48427685 - 6959}{2}) + 2 πn

Show solutions in decimal form

θ = 1.57065 \dots + 2 πn, θ = 2 π - 1.57065 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 1-cos^2(θ)-6959cos(θ)=0 ?
The general solution for 1-cos^2(θ)-6959cos(θ)=0 is θ=1.57065…+2pin,θ=2pi-1.57065…+2pin