What is the general solution for cos^2(θ)+(20^2cos(θ))/(9.81(1))-1=0 ?

The general solution for cos^2(θ)+(20^2cos(θ))/(9.81(1))-1=0 is θ=1.54628…+2pin,θ=2pi-1.54628…+2pin

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cos^2(θ)+(20^2cos(θ))/(9.81(1))-1=0

Solution

cos^{2} (θ) + \frac{2 0 ^{2} cos ( θ )}{9.81 ( 1 )} - 1 = 0

Solution

θ = 1.54628 \dots + 2 πn, θ = 2 π - 1.54628 \dots + 2 πn

Degrees

θ = 88.59552 \dots^{\circ} + 36 0^{\circ} n, θ = 271.40447 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (θ) + \frac{2 0 ^{2} cos ( θ )}{9.81 ( 1 )} - 1 = 0

Solve by substitution

cos^{2} (θ) + \frac{2 0 ^{2} cos ( θ )}{9.81 \cdot 1} - 1 = 0

Let:

cos (θ) = u

u^{2} + \frac{2 0 ^{2} u}{9.81 \cdot 1} - 1 = 0

u^{2} + \frac{2 0 ^{2} u}{9.81 \cdot 1} - 1 = 0 : u = \frac{- 400 + 160384.9444}{19.62}, u = \frac{- 400 - 160384.9444}{19.62}

u^{2} + \frac{2 0 ^{2} u}{9.81 \cdot 1} - 1 = 0

Simplify

9.81 \cdot 1 : 9.81

9.81 \cdot 1

Multiply the numbers:

9.81 \cdot 1 = 9.81

= 9.81

u^{2} + \frac{2 0 ^{2} u}{9.81} - 1 = 0

Multiply both sides by

9.81

u^{2} + \frac{2 0 ^{2} u}{9.81} - 1 = 0

Multiply both sides by

9.81

u^{2} \cdot 9.81 + \frac{2 0 ^{2} u}{9.81} \cdot 9.81 - 1 \cdot 9.81 = 0 \cdot 9.81

Simplify

9.81 u^{2} + 400 u - 9.81 = 0

9.81 u^{2} + 400 u - 9.81 = 0

Solve with the quadratic formula

9.81 u^{2} + 400 u - 9.81 = 0

Quadratic Equation Formula:

For

a = 9.81, b = 400, c = - 9.81

u_{1, 2} = \frac{- 400 \pm 40 0 ^{2} - 4 \cdot 9.81 ( - 9.81 )}{2 \cdot 9.81}

u_{1, 2} = \frac{- 400 \pm 40 0 ^{2} - 4 \cdot 9.81 ( - 9.81 )}{2 \cdot 9.81}

40 0^{2} - 4 \cdot 9.81 (- 9.81) = 160384.9444

40 0^{2} - 4 \cdot 9.81 (- 9.81)

Apply rule

- (- a) = a

= 40 0^{2} + 4 \cdot 9.81 \cdot 9.81

Multiply the numbers:

4 \cdot 9.81 \cdot 9.81 = 384.9444

= 40 0^{2} + 384.9444

40 0^{2} = 160000

= 160000 + 384.9444

Add the numbers:

160000 + 384.9444 = 160384.9444

= 160384.9444

u_{1, 2} = \frac{- 400 \pm 160384.9444}{2 \cdot 9.81}

Separate the solutions

u_{1} = \frac{- 400 + 160384.9444}{2 \cdot 9.81}, u_{2} = \frac{- 400 - 160384.9444}{2 \cdot 9.81}

u = \frac{- 400 + 160384.9444}{2 \cdot 9.81} : \frac{- 400 + 160384.9444}{19.62}

\frac{- 400 + 160384.9444}{2 \cdot 9.81}

Multiply the numbers:

2 \cdot 9.81 = 19.62

= \frac{- 400 + 160384.9444}{19.62}

u = \frac{- 400 - 160384.9444}{2 \cdot 9.81} : \frac{- 400 - 160384.9444}{19.62}

\frac{- 400 - 160384.9444}{2 \cdot 9.81}

Multiply the numbers:

2 \cdot 9.81 = 19.62

= \frac{- 400 - 160384.9444}{19.62}

The solutions to the quadratic equation are:

u = \frac{- 400 + 160384.9444}{19.62}, u = \frac{- 400 - 160384.9444}{19.62}

Substitute back

u = cos (θ)

cos (θ) = \frac{- 400 + 160384.9444}{19.62}, cos (θ) = \frac{- 400 - 160384.9444}{19.62}

cos (θ) = \frac{- 400 + 160384.9444}{19.62}, cos (θ) = \frac{- 400 - 160384.9444}{19.62}

cos (θ) = \frac{- 400 + 160384.9444}{19.62} : θ = arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn, θ = 2 π - arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn

cos (θ) = \frac{- 400 + 160384.9444}{19.62}

Apply trig inverse properties

cos (θ) = \frac{- 400 + 160384.9444}{19.62}

General solutions for

cos (θ) = \frac{- 400 + 160384.9444}{19.62}

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

θ = arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn, θ = 2 π - arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn

θ = arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn, θ = 2 π - arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn

cos (θ) = \frac{- 400 - 160384.9444}{19.62} :

No Solution

cos (θ) = \frac{- 400 - 160384.9444}{19.62}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn, θ = 2 π - arccos (\frac{- 400 + 160384.9444}{19.62}) + 2 πn

Show solutions in decimal form

θ = 1.54628 \dots + 2 πn, θ = 2 π - 1.54628 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(θ)+(20^2cos(θ))/(9.81(1))-1=0 ?
The general solution for cos^2(θ)+(20^2cos(θ))/(9.81(1))-1=0 is θ=1.54628…+2pin,θ=2pi-1.54628…+2pin