What is the general solution for cos^2(θ)+(5^2cos(θ))/(9.81(8))-1=0 ?

The general solution for cos^2(θ)+(5^2cos(θ))/(9.81(8))-1=0 is θ=0.54845…+2pin,θ=2pi-0.54845…+2pin

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Popular >

cos^2(θ)+(5^2cos(θ))/(9.81(8))-1=0

Solution

cos^{2} (θ) + \frac{5 ^{2} cos ( θ )}{9.81 ( 8 )} - 1 = 0

Solution

θ = 0.54845 \dots + 2 πn, θ = 2 π - 0.54845 \dots + 2 πn

Degrees

θ = 31.42440 \dots^{\circ} + 36 0^{\circ} n, θ = 328.57559 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (θ) + \frac{5 ^{2} cos ( θ )}{9.81 ( 8 )} - 1 = 0

Solve by substitution

cos^{2} (θ) + \frac{5 ^{2} cos ( θ )}{9.81 \cdot 8} - 1 = 0

Let:

cos (θ) = u

u^{2} + \frac{5 ^{2} u}{9.81 \cdot 8} - 1 = 0

u^{2} + \frac{5 ^{2} u}{9.81 \cdot 8} - 1 = 0 : u = \frac{- 25 + 25261.4416}{156.96}, u = \frac{- 25 - 25261.4416}{156.96}

u^{2} + \frac{5 ^{2} u}{9.81 \cdot 8} - 1 = 0

Simplify

9.81 \cdot 8 : 78.48

9.81 \cdot 8

Multiply the numbers:

9.81 \cdot 8 = 78.48

= 78.48

u^{2} + \frac{5 ^{2} u}{78.48} - 1 = 0

Multiply both sides by

78.48

u^{2} + \frac{5 ^{2} u}{78.48} - 1 = 0

Multiply both sides by

78.48

u^{2} \cdot 78.48 + \frac{5 ^{2} u}{78.48} \cdot 78.48 - 1 \cdot 78.48 = 0 \cdot 78.48

Simplify

78.48 u^{2} + 25 u - 78.48 = 0

78.48 u^{2} + 25 u - 78.48 = 0

Solve with the quadratic formula

78.48 u^{2} + 25 u - 78.48 = 0

Quadratic Equation Formula:

For

a = 78.48, b = 25, c = - 78.48

u_{1, 2} = \frac{- 25 \pm 2 5 ^{2} - 4 \cdot 78.48 ( - 78.48 )}{2 \cdot 78.48}

u_{1, 2} = \frac{- 25 \pm 2 5 ^{2} - 4 \cdot 78.48 ( - 78.48 )}{2 \cdot 78.48}

2 5^{2} - 4 \cdot 78.48 (- 78.48) = 25261.4416

2 5^{2} - 4 \cdot 78.48 (- 78.48)

Apply rule

- (- a) = a

= 2 5^{2} + 4 \cdot 78.48 \cdot 78.48

Multiply the numbers:

4 \cdot 78.48 \cdot 78.48 = 24636.4416

= 2 5^{2} + 24636.4416

2 5^{2} = 625

= 625 + 24636.4416

Add the numbers:

625 + 24636.4416 = 25261.4416

= 25261.4416

u_{1, 2} = \frac{- 25 \pm 25261.4416}{2 \cdot 78.48}

Separate the solutions

u_{1} = \frac{- 25 + 25261.4416}{2 \cdot 78.48}, u_{2} = \frac{- 25 - 25261.4416}{2 \cdot 78.48}

u = \frac{- 25 + 25261.4416}{2 \cdot 78.48} : \frac{- 25 + 25261.4416}{156.96}

\frac{- 25 + 25261.4416}{2 \cdot 78.48}

Multiply the numbers:

2 \cdot 78.48 = 156.96

= \frac{- 25 + 25261.4416}{156.96}

u = \frac{- 25 - 25261.4416}{2 \cdot 78.48} : \frac{- 25 - 25261.4416}{156.96}

\frac{- 25 - 25261.4416}{2 \cdot 78.48}

Multiply the numbers:

2 \cdot 78.48 = 156.96

= \frac{- 25 - 25261.4416}{156.96}

The solutions to the quadratic equation are:

u = \frac{- 25 + 25261.4416}{156.96}, u = \frac{- 25 - 25261.4416}{156.96}

Substitute back

u = cos (θ)

cos (θ) = \frac{- 25 + 25261.4416}{156.96}, cos (θ) = \frac{- 25 - 25261.4416}{156.96}

cos (θ) = \frac{- 25 + 25261.4416}{156.96}, cos (θ) = \frac{- 25 - 25261.4416}{156.96}

cos (θ) = \frac{- 25 + 25261.4416}{156.96} : θ = arccos (\frac{- 25 + 25261.4416}{156.96}) + 2 πn, θ = 2 π - arccos (\frac{- 25 + 25261.4416}{156.96}) + 2 πn

cos (θ) = \frac{- 25 + 25261.4416}{156.96}

Apply trig inverse properties

cos (θ) = \frac{- 25 + 25261.4416}{156.96}

General solutions for

cos (θ) = \frac{- 25 + 25261.4416}{156.96}

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

θ = arccos (\frac{- 25 + 25261.4416}{156.96}) + 2 πn, θ = 2 π - arccos (\frac{- 25 + 25261.4416}{156.96}) + 2 πn

θ = arccos (\frac{- 25 + 25261.4416}{156.96}) + 2 πn, θ = 2 π - arccos (\frac{- 25 + 25261.4416}{156.96}) + 2 πn

cos (θ) = \frac{- 25 - 25261.4416}{156.96} :

No Solution

cos (θ) = \frac{- 25 - 25261.4416}{156.96}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = arccos (\frac{- 25 + 25261.4416}{156.96}) + 2 πn, θ = 2 π - arccos (\frac{- 25 + 25261.4416}{156.96}) + 2 πn

Show solutions in decimal form

θ = 0.54845 \dots + 2 πn, θ = 2 π - 0.54845 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(θ)+(5^2cos(θ))/(9.81(8))-1=0 ?
The general solution for cos^2(θ)+(5^2cos(θ))/(9.81(8))-1=0 is θ=0.54845…+2pin,θ=2pi-0.54845…+2pin