What is the general solution for solvefor a,-10=sin(a)+15.2cos(a) ?

The general solution for solvefor a,-10=sin(a)+15.2cos(a) is a=2.35262…+2pin,a=-2.22123…+2pin

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Popular >

solvefor a,-10=sin(a)+15.2cos(a)

Solution

solve for

a, - 10 = sin (a) + 15.2 cos (a)

Solution

a = 2.35262 \dots + 2 πn, a = - 2.22123 \dots + 2 πn

Degrees

a = 134.79566 \dots^{\circ} + 36 0^{\circ} n, a = - 127.26759 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 10 = sin (a) + 15.2 cos (a)

Subtract

15.2 cos (a)

from both sides

sin (a) = - 10 - 15.2 cos (a)

Square both sides

sin^{2} (a) = (- 10 - 15.2 cos (a))^{2}

Subtract

(- 10 - 15.2 cos (a))^{2}

from both sides

sin^{2} (a) - 100 - 304 cos (a) - 231.04 cos^{2} (a) = 0

Rewrite using trig identities

- 100 + sin^{2} (a) - 231.04 cos^{2} (a) - 304 cos (a)

Use the Pythagorean identity:

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

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

= - 100 + 1 - cos^{2} (a) - 231.04 cos^{2} (a) - 304 cos (a)

Simplify

- 100 + 1 - cos^{2} (a) - 231.04 cos^{2} (a) - 304 cos (a) : - 232.04 cos^{2} (a) - 304 cos (a) - 99

- 100 + 1 - cos^{2} (a) - 231.04 cos^{2} (a) - 304 cos (a)

Add similar elements:

- cos^{2} (a) - 231.04 cos^{2} (a) = - 232.04 cos^{2} (a)

= - 100 + 1 - 232.04 cos^{2} (a) - 304 cos (a)

Add/Subtract the numbers:

- 100 + 1 = - 99

= - 232.04 cos^{2} (a) - 304 cos (a) - 99

= - 232.04 cos^{2} (a) - 304 cos (a) - 99

- 99 - 232.04 cos^{2} (a) - 304 cos (a) = 0

Solve by substitution

- 99 - 232.04 cos^{2} (a) - 304 cos (a) = 0

Let:

cos (a) = u

- 99 - 232.04 u^{2} - 304 u = 0

- 99 - 232.04 u^{2} - 304 u = 0 : u = - \frac{5 ( 760 + 3301 )}{5801}, u = - \frac{5 ( 760 - 3301 )}{5801}

- 99 - 232.04 u^{2} - 304 u = 0

Multiply both sides by

100

- 99 - 232.04 u^{2} - 304 u = 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

- 99 \cdot 100 - 232.04 u^{2} \cdot 100 - 304 u \cdot 100 = 0 \cdot 100

Refine

- 9900 - 23204 u^{2} - 30400 u = 0

- 9900 - 23204 u^{2} - 30400 u = 0

Write in the standard form

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

- 23204 u^{2} - 30400 u - 9900 = 0

Solve with the quadratic formula

- 23204 u^{2} - 30400 u - 9900 = 0

Quadratic Equation Formula:

For

a = - 23204, b = - 30400, c = - 9900

u_{1, 2} = \frac{- ( - 30400 ) \pm ( - 30400 ) ^{2} - 4 ( - 23204 ) ( - 9900 )}{2 ( - 23204 )}

u_{1, 2} = \frac{- ( - 30400 ) \pm ( - 30400 ) ^{2} - 4 ( - 23204 ) ( - 9900 )}{2 ( - 23204 )}

(- 30400)^{2} - 4 (- 23204) (- 9900) = 403301

(- 30400)^{2} - 4 (- 23204) (- 9900)

Apply rule

- (- a) = a

= (- 30400)^{2} - 4 \cdot 23204 \cdot 9900

Apply exponent rule:

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

n

is even

(- 30400)^{2} = 3040 0^{2}

= 3040 0^{2} - 4 \cdot 23204 \cdot 9900

Multiply the numbers:

4 \cdot 23204 \cdot 9900 = 918878400

= 3040 0^{2} - 918878400

3040 0^{2} = 924160000

= 924160000 - 918878400

Subtract the numbers:

924160000 - 918878400 = 5281600

= 5281600

Prime factorization of

5281600 : 2^{6} \cdot 5^{2} \cdot 3301

5281600

= 2^{6} \cdot 5^{2} \cdot 3301

Apply radical rule:

= 3301 2^{6} 5^{2}

Apply radical rule:

2^{6} = 2^{\frac{6}{2}} = 2^{3}

= 2^{3} 3301 5^{2}

Apply radical rule:

5^{2} = 5

= 2^{3} \cdot 53301

Refine

= 403301

u_{1, 2} = \frac{- ( - 30400 ) \pm 40 3301}{2 ( - 23204 )}

Separate the solutions

u_{1} = \frac{- ( - 30400 ) + 40 3301}{2 ( - 23204 )}, u_{2} = \frac{- ( - 30400 ) - 40 3301}{2 ( - 23204 )}

u = \frac{- ( - 30400 ) + 40 3301}{2 ( - 23204 )} : - \frac{5 ( 760 + 3301 )}{5801}

\frac{- ( - 30400 ) + 40 3301}{2 ( - 23204 )}

Remove parentheses:

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

= \frac{30400 + 40 3301}{- 2 \cdot 23204}

Multiply the numbers:

2 \cdot 23204 = 46408

= \frac{30400 + 40 3301}{- 46408}

Apply the fraction rule:

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

= - \frac{30400 + 40 3301}{46408}

Cancel

\frac{30400 + 40 3301}{46408} : \frac{5 ( 760 + 3301 )}{5801}

\frac{30400 + 40 3301}{46408}

Factor

30400 + 403301 : 40 (760 + 3301)

30400 + 403301

Rewrite as

= 40 \cdot 760 + 403301

Factor out common term

40

= 40 (760 + 3301)

= \frac{40 ( 760 + 3301 )}{46408}

Cancel the common factor:

8

= \frac{5 ( 760 + 3301 )}{5801}

= - \frac{5 ( 760 + 3301 )}{5801}

u = \frac{- ( - 30400 ) - 40 3301}{2 ( - 23204 )} : - \frac{5 ( 760 - 3301 )}{5801}

\frac{- ( - 30400 ) - 40 3301}{2 ( - 23204 )}

Remove parentheses:

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

= \frac{30400 - 40 3301}{- 2 \cdot 23204}

Multiply the numbers:

2 \cdot 23204 = 46408

= \frac{30400 - 40 3301}{- 46408}

Apply the fraction rule:

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

= - \frac{30400 - 40 3301}{46408}

Cancel

\frac{30400 - 40 3301}{46408} : \frac{5 ( 760 - 3301 )}{5801}

\frac{30400 - 40 3301}{46408}

Factor

30400 - 403301 : 40 (760 - 3301)

30400 - 403301

Rewrite as

= 40 \cdot 760 - 403301

Factor out common term

40

= 40 (760 - 3301)

= \frac{40 ( 760 - 3301 )}{46408}

Cancel the common factor:

8

= \frac{5 ( 760 - 3301 )}{5801}

= - \frac{5 ( 760 - 3301 )}{5801}

The solutions to the quadratic equation are:

u = - \frac{5 ( 760 + 3301 )}{5801}, u = - \frac{5 ( 760 - 3301 )}{5801}

Substitute back

u = cos (a)

cos (a) = - \frac{5 ( 760 + 3301 )}{5801}, cos (a) = - \frac{5 ( 760 - 3301 )}{5801}

cos (a) = - \frac{5 ( 760 + 3301 )}{5801}, cos (a) = - \frac{5 ( 760 - 3301 )}{5801}

cos (a) = - \frac{5 ( 760 + 3301 )}{5801} : a = arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn, a = - arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn

cos (a) = - \frac{5 ( 760 + 3301 )}{5801}

Apply trig inverse properties

cos (a) = - \frac{5 ( 760 + 3301 )}{5801}

General solutions for

cos (a) = - \frac{5 ( 760 + 3301 )}{5801}

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

a = arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn, a = - arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn

a = arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn, a = - arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn

cos (a) = - \frac{5 ( 760 - 3301 )}{5801} : a = arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn, a = - arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn

cos (a) = - \frac{5 ( 760 - 3301 )}{5801}

Apply trig inverse properties

cos (a) = - \frac{5 ( 760 - 3301 )}{5801}

General solutions for

cos (a) = - \frac{5 ( 760 - 3301 )}{5801}

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

a = arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn, a = - arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn

a = arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn, a = - arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn

Combine all the solutions

a = arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn, a = - arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn, a = arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn, a = - arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sin (a) + 15.2 cos (a) = - 10

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

Check the solution

arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn :

True

arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn

Plug in

n = 1

arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 π 1

For

sin (a) + 15.2 cos (a) = - 10

plug in

a = arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 π 1

sin (arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 π 1) + 15.2 cos (arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 π 1) = - 10

Refine

- 10 = - 10

\Rightarrow True

Check the solution

- arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn :

False

- arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn

Plug in

n = 1

- arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 π 1

For

sin (a) + 15.2 cos (a) = - 10

plug in

a = - arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 π 1

sin (- arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 π 1) + 15.2 cos (- arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 π 1) = - 10

Refine

- 11.41924 \dots = - 10

\Rightarrow False

Check the solution

arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn :

False

arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn

Plug in

n = 1

arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 π 1

For

sin (a) + 15.2 cos (a) = - 10

plug in

a = arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 π 1

sin (arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 π 1) + 15.2 cos (arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 π 1) = - 10

Refine

- 8.40836 \dots = - 10

\Rightarrow False

Check the solution

- arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn :

True

- arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn

Plug in

n = 1

- arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 π 1

For

sin (a) + 15.2 cos (a) = - 10

plug in

a = - arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 π 1

sin (- arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 π 1) + 15.2 cos (- arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 π 1) = - 10

Refine

- 10 = - 10

\Rightarrow True

a = arccos (- \frac{5 ( 760 + 3301 )}{5801}) + 2 πn, a = - arccos (- \frac{5 ( 760 - 3301 )}{5801}) + 2 πn

Show solutions in decimal form

a = 2.35262 \dots + 2 πn, a = - 2.22123 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor a,-10=sin(a)+15.2cos(a) ?
The general solution for solvefor a,-10=sin(a)+15.2cos(a) is a=2.35262…+2pin,a=-2.22123…+2pin