What is the general solution for 25sin(θ)-1.5cos(θ)=20 ?

The general solution for 25sin(θ)-1.5cos(θ)=20 is θ=2.27661…+2pin,θ=0.98483…+2pin

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25sin(θ)-1.5cos(θ)=20

Solution

25 sin (θ) - 1.5 cos (θ) = 20

Solution

θ = 2.27661 \dots + 2 πn, θ = 0.98483 \dots + 2 πn

Degrees

θ = 130.44044 \dots^{\circ} + 36 0^{\circ} n, θ = 56.42681 \dots^{\circ} + 36 0^{\circ} n

Solution steps

25 sin (θ) - 1.5 cos (θ) = 20

Add

1.5 cos (θ)

to both sides

25 sin (θ) = 20 + 1.5 cos (θ)

Square both sides

(25 sin (θ))^{2} = (20 + 1.5 cos (θ))^{2}

Subtract

(20 + 1.5 cos (θ))^{2}

from both sides

625 sin^{2} (θ) - 400 - 60 cos (θ) - 2.25 cos^{2} (θ) = 0

Rewrite using trig identities

- 400 - 2.25 cos^{2} (θ) - 60 cos (θ) + 625 sin^{2} (θ)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

= - 400 - 2.25 cos^{2} (θ) - 60 cos (θ) + 625 (1 - cos^{2} (θ))

Simplify

- 400 - 2.25 cos^{2} (θ) - 60 cos (θ) + 625 (1 - cos^{2} (θ)) : - 627.25 cos^{2} (θ) - 60 cos (θ) + 225

- 400 - 2.25 cos^{2} (θ) - 60 cos (θ) + 625 (1 - cos^{2} (θ))

Expand

625 (1 - cos^{2} (θ)) : 625 - 625 cos^{2} (θ)

625 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 625, b = 1, c = cos^{2} (θ)

= 625 \cdot 1 - 625 cos^{2} (θ)

Multiply the numbers:

625 \cdot 1 = 625

= 625 - 625 cos^{2} (θ)

= - 400 - 2.25 cos^{2} (θ) - 60 cos (θ) + 625 - 625 cos^{2} (θ)

Simplify

- 400 - 2.25 cos^{2} (θ) - 60 cos (θ) + 625 - 625 cos^{2} (θ) : - 627.25 cos^{2} (θ) - 60 cos (θ) + 225

- 400 - 2.25 cos^{2} (θ) - 60 cos (θ) + 625 - 625 cos^{2} (θ)

Group like terms

= - 2.25 cos^{2} (θ) - 60 cos (θ) - 625 cos^{2} (θ) - 400 + 625

Add similar elements:

- 2.25 cos^{2} (θ) - 625 cos^{2} (θ) = - 627.25 cos^{2} (θ)

= - 627.25 cos^{2} (θ) - 60 cos (θ) - 400 + 625

Add/Subtract the numbers:

- 400 + 625 = 225

= - 627.25 cos^{2} (θ) - 60 cos (θ) + 225

= - 627.25 cos^{2} (θ) - 60 cos (θ) + 225

= - 627.25 cos^{2} (θ) - 60 cos (θ) + 225

225 - 60 cos (θ) - 627.25 cos^{2} (θ) = 0

Solve by substitution

225 - 60 cos (θ) - 627.25 cos^{2} (θ) = 0

Let:

cos (θ) = u

225 - 60 u - 627.25 u^{2} = 0

225 - 60 u - 627.25 u^{2} = 0 : u = - \frac{6000 + 5681250000}{125450}, u = \frac{5681250000 - 6000}{125450}

225 - 60 u - 627.25 u^{2} = 0

Multiply both sides by

100

225 - 60 u - 627.25 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

225 \cdot 100 - 60 u \cdot 100 - 627.25 u^{2} \cdot 100 = 0 \cdot 100

Refine

22500 - 6000 u - 62725 u^{2} = 0

22500 - 6000 u - 62725 u^{2} = 0

Write in the standard form

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

- 62725 u^{2} - 6000 u + 22500 = 0

Solve with the quadratic formula

- 62725 u^{2} - 6000 u + 22500 = 0

Quadratic Equation Formula:

For

a = - 62725, b = - 6000, c = 22500

u_{1, 2} = \frac{- ( - 6000 ) \pm ( - 6000 ) ^{2} - 4 ( - 62725 ) \cdot 22500}{2 ( - 62725 )}

u_{1, 2} = \frac{- ( - 6000 ) \pm ( - 6000 ) ^{2} - 4 ( - 62725 ) \cdot 22500}{2 ( - 62725 )}

(- 6000)^{2} - 4 (- 62725) \cdot 22500 = 5681250000

(- 6000)^{2} - 4 (- 62725) \cdot 22500

Apply rule

- (- a) = a

= (- 6000)^{2} + 4 \cdot 62725 \cdot 22500

Apply exponent rule:

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

n

is even

(- 6000)^{2} = 600 0^{2}

= 600 0^{2} + 4 \cdot 62725 \cdot 22500

Multiply the numbers:

4 \cdot 62725 \cdot 22500 = 5645250000

= 600 0^{2} + 5645250000

600 0^{2} = 36000000

= 36000000 + 5645250000

Add the numbers:

36000000 + 5645250000 = 5681250000

= 5681250000

u_{1, 2} = \frac{- ( - 6000 ) \pm 5681250000}{2 ( - 62725 )}

Separate the solutions

u_{1} = \frac{- ( - 6000 ) + 5681250000}{2 ( - 62725 )}, u_{2} = \frac{- ( - 6000 ) - 5681250000}{2 ( - 62725 )}

u = \frac{- ( - 6000 ) + 5681250000}{2 ( - 62725 )} : - \frac{6000 + 5681250000}{125450}

\frac{- ( - 6000 ) + 5681250000}{2 ( - 62725 )}

Remove parentheses:

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

= \frac{6000 + 5681250000}{- 2 \cdot 62725}

Multiply the numbers:

2 \cdot 62725 = 125450

= \frac{6000 + 5681250000}{- 125450}

Apply the fraction rule:

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

= - \frac{6000 + 5681250000}{125450}

u = \frac{- ( - 6000 ) - 5681250000}{2 ( - 62725 )} : \frac{5681250000 - 6000}{125450}

\frac{- ( - 6000 ) - 5681250000}{2 ( - 62725 )}

Remove parentheses:

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

= \frac{6000 - 5681250000}{- 2 \cdot 62725}

Multiply the numbers:

2 \cdot 62725 = 125450

= \frac{6000 - 5681250000}{- 125450}

Apply the fraction rule:

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

6000 - 5681250000 = - (5681250000 - 6000)

= \frac{5681250000 - 6000}{125450}

The solutions to the quadratic equation are:

u = - \frac{6000 + 5681250000}{125450}, u = \frac{5681250000 - 6000}{125450}

Substitute back

u = cos (θ)

cos (θ) = - \frac{6000 + 5681250000}{125450}, cos (θ) = \frac{5681250000 - 6000}{125450}

cos (θ) = - \frac{6000 + 5681250000}{125450}, cos (θ) = \frac{5681250000 - 6000}{125450}

cos (θ) = - \frac{6000 + 5681250000}{125450} : θ = arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn, θ = - arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn

cos (θ) = - \frac{6000 + 5681250000}{125450}

Apply trig inverse properties

cos (θ) = - \frac{6000 + 5681250000}{125450}

General solutions for

cos (θ) = - \frac{6000 + 5681250000}{125450}

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

θ = arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn, θ = - arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn

θ = arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn, θ = - arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn

cos (θ) = \frac{5681250000 - 6000}{125450} : θ = arccos (\frac{5681250000 - 6000}{125450}) + 2 πn, θ = 2 π - arccos (\frac{5681250000 - 6000}{125450}) + 2 πn

cos (θ) = \frac{5681250000 - 6000}{125450}

Apply trig inverse properties

cos (θ) = \frac{5681250000 - 6000}{125450}

General solutions for

cos (θ) = \frac{5681250000 - 6000}{125450}

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

θ = arccos (\frac{5681250000 - 6000}{125450}) + 2 πn, θ = 2 π - arccos (\frac{5681250000 - 6000}{125450}) + 2 πn

θ = arccos (\frac{5681250000 - 6000}{125450}) + 2 πn, θ = 2 π - arccos (\frac{5681250000 - 6000}{125450}) + 2 πn

Combine all the solutions

θ = arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn, θ = - arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn, θ = arccos (\frac{5681250000 - 6000}{125450}) + 2 πn, θ = 2 π - arccos (\frac{5681250000 - 6000}{125450}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

25 sin (θ) - 1.5 cos (θ) = 20

Remove the ones that don't agree with the equation.

Check the solution

arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn :

True

arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn

Plug in

n = 1

arccos (- \frac{6000 + 5681250000}{125450}) + 2 π 1

For

25 sin (θ) - 1.5 cos (θ) = 20

plug in

θ = arccos (- \frac{6000 + 5681250000}{125450}) + 2 π 1

25 sin (arccos (- \frac{6000 + 5681250000}{125450}) + 2 π 1) - 1.5 cos (arccos (- \frac{6000 + 5681250000}{125450}) + 2 π 1) = 20

Refine

20 = 20

\Rightarrow True

Check the solution

- arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn :

False

- arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn

Plug in

n = 1

- arccos (- \frac{6000 + 5681250000}{125450}) + 2 π 1

For

25 sin (θ) - 1.5 cos (θ) = 20

plug in

θ = - arccos (- \frac{6000 + 5681250000}{125450}) + 2 π 1

25 sin (- arccos (- \frac{6000 + 5681250000}{125450}) + 2 π 1) - 1.5 cos (- arccos (- \frac{6000 + 5681250000}{125450}) + 2 π 1) = 20

Refine

- 18.05402 \dots = 20

\Rightarrow False

Check the solution

arccos (\frac{5681250000 - 6000}{125450}) + 2 πn :

True

arccos (\frac{5681250000 - 6000}{125450}) + 2 πn

Plug in

n = 1

arccos (\frac{5681250000 - 6000}{125450}) + 2 π 1

For

25 sin (θ) - 1.5 cos (θ) = 20

plug in

θ = arccos (\frac{5681250000 - 6000}{125450}) + 2 π 1

25 sin (arccos (\frac{5681250000 - 6000}{125450}) + 2 π 1) - 1.5 cos (arccos (\frac{5681250000 - 6000}{125450}) + 2 π 1) = 20

Refine

20 = 20

\Rightarrow True

Check the solution

2 π - arccos (\frac{5681250000 - 6000}{125450}) + 2 πn :

False

2 π - arccos (\frac{5681250000 - 6000}{125450}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{5681250000 - 6000}{125450}) + 2 π 1

For

25 sin (θ) - 1.5 cos (θ) = 20

plug in

θ = 2 π - arccos (\frac{5681250000 - 6000}{125450}) + 2 π 1

25 sin (2 π - arccos (\frac{5681250000 - 6000}{125450}) + 2 π 1) - 1.5 cos (2 π - arccos (\frac{5681250000 - 6000}{125450}) + 2 π 1) = 20

Refine

- 21.65900 \dots = 20

\Rightarrow False

θ = arccos (- \frac{6000 + 5681250000}{125450}) + 2 πn, θ = arccos (\frac{5681250000 - 6000}{125450}) + 2 πn

Show solutions in decimal form

θ = 2.27661 \dots + 2 πn, θ = 0.98483 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 25sin(θ)-1.5cos(θ)=20 ?
The general solution for 25sin(θ)-1.5cos(θ)=20 is θ=2.27661…+2pin,θ=0.98483…+2pin