What is the general solution for 15sin(x)+6cos(x)-3=0 ?

The general solution for 15sin(x)+6cos(x)-3=0 is x=pi-0.56728…+2pin,x=-0.19372…+2pin

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15sin(x)+6cos(x)-3=0

Solution

15 sin (x) + 6 cos (x) - 3 = 0

Solution

x = π - 0.56728 \dots + 2 πn, x = - 0.19372 \dots + 2 πn

Degrees

x = 147.49691 \dots^{\circ} + 36 0^{\circ} n, x = - 11.09973 \dots^{\circ} + 36 0^{\circ} n

Solution steps

15 sin (x) + 6 cos (x) - 3 = 0

Subtract

6 cos (x)

from both sides

15 sin (x) - 3 = - 6 cos (x)

Square both sides

(15 sin (x) - 3)^{2} = (- 6 cos (x))^{2}

Subtract

(- 6 cos (x))^{2}

from both sides

(15 sin (x) - 3)^{2} - 36 cos^{2} (x) = 0

Rewrite using trig identities

(- 3 + 15 sin (x))^{2} - 36 cos^{2} (x)

Use the Pythagorean identity:

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

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

= (- 3 + 15 sin (x))^{2} - 36 (1 - sin^{2} (x))

Simplify

(- 3 + 15 sin (x))^{2} - 36 (1 - sin^{2} (x)) : 261 sin^{2} (x) - 90 sin (x) - 27

(- 3 + 15 sin (x))^{2} - 36 (1 - sin^{2} (x))

(- 3 + 15 sin (x))^{2} : 9 - 90 sin (x) + 225 sin^{2} (x)

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = - 3, b = 15 sin (x)

= (- 3)^{2} + 2 (- 3) \cdot 15 sin (x) + (15 sin (x))^{2}

Simplify

(- 3)^{2} + 2 (- 3) \cdot 15 sin (x) + (15 sin (x))^{2} : 9 - 90 sin (x) + 225 sin^{2} (x)

(- 3)^{2} + 2 (- 3) \cdot 15 sin (x) + (15 sin (x))^{2}

Remove parentheses:

(- a) = - a

= (- 3)^{2} - 2 \cdot 3 \cdot 15 sin (x) + (15 sin (x))^{2}

(- 3)^{2} = 9

(- 3)^{2}

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

= 3^{2}

3^{2} = 9

= 9

2 \cdot 3 \cdot 15 sin (x) = 90 sin (x)

2 \cdot 3 \cdot 15 sin (x)

Multiply the numbers:

2 \cdot 3 \cdot 15 = 90

= 90 sin (x)

(15 sin (x))^{2} = 225 sin^{2} (x)

(15 sin (x))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 1 5^{2} sin^{2} (x)

1 5^{2} = 225

= 225 sin^{2} (x)

= 9 - 90 sin (x) + 225 sin^{2} (x)

= 9 - 90 sin (x) + 225 sin^{2} (x)

= 9 - 90 sin (x) + 225 sin^{2} (x) - 36 (1 - sin^{2} (x))

Expand

- 36 (1 - sin^{2} (x)) : - 36 + 36 sin^{2} (x)

- 36 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 36, b = 1, c = sin^{2} (x)

= - 36 \cdot 1 - (- 36) sin^{2} (x)

Apply minus-plus rules

- (- a) = a

= - 36 \cdot 1 + 36 sin^{2} (x)

Multiply the numbers:

36 \cdot 1 = 36

= - 36 + 36 sin^{2} (x)

= 9 - 90 sin (x) + 225 sin^{2} (x) - 36 + 36 sin^{2} (x)

Simplify

9 - 90 sin (x) + 225 sin^{2} (x) - 36 + 36 sin^{2} (x) : 261 sin^{2} (x) - 90 sin (x) - 27

9 - 90 sin (x) + 225 sin^{2} (x) - 36 + 36 sin^{2} (x)

Group like terms

= - 90 sin (x) + 225 sin^{2} (x) + 36 sin^{2} (x) + 9 - 36

Add similar elements:

225 sin^{2} (x) + 36 sin^{2} (x) = 261 sin^{2} (x)

= - 90 sin (x) + 261 sin^{2} (x) + 9 - 36

Add/Subtract the numbers:

9 - 36 = - 27

= 261 sin^{2} (x) - 90 sin (x) - 27

= 261 sin^{2} (x) - 90 sin (x) - 27

= 261 sin^{2} (x) - 90 sin (x) - 27

- 27 + 261 sin^{2} (x) - 90 sin (x) = 0

Solve by substitution

- 27 + 261 sin^{2} (x) - 90 sin (x) = 0

Let:

sin (x) = u

- 27 + 261 u^{2} - 90 u = 0

- 27 + 261 u^{2} - 90 u = 0 : u = \frac{5 + 4 7}{29}, u = \frac{5 - 4 7}{29}

- 27 + 261 u^{2} - 90 u = 0

Write in the standard form

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

261 u^{2} - 90 u - 27 = 0

Solve with the quadratic formula

261 u^{2} - 90 u - 27 = 0

Quadratic Equation Formula:

For

a = 261, b = - 90, c = - 27

u_{1, 2} = \frac{- ( - 90 ) \pm ( - 90 ) ^{2} - 4 \cdot 261 ( - 27 )}{2 \cdot 261}

u_{1, 2} = \frac{- ( - 90 ) \pm ( - 90 ) ^{2} - 4 \cdot 261 ( - 27 )}{2 \cdot 261}

(- 90)^{2} - 4 \cdot 261 (- 27) = 727

(- 90)^{2} - 4 \cdot 261 (- 27)

Apply rule

- (- a) = a

= (- 90)^{2} + 4 \cdot 261 \cdot 27

Apply exponent rule:

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

n

is even

(- 90)^{2} = 9 0^{2}

= 9 0^{2} + 4 \cdot 261 \cdot 27

Multiply the numbers:

4 \cdot 261 \cdot 27 = 28188

= 9 0^{2} + 28188

9 0^{2} = 8100

= 8100 + 28188

Add the numbers:

8100 + 28188 = 36288

= 36288

Prime factorization of

36288 : 2^{6} \cdot 3^{4} \cdot 7

36288

36288

divides by

236288 = 18144 \cdot 2

= 2 \cdot 18144

18144

divides by

218144 = 9072 \cdot 2

= 2 \cdot 2 \cdot 9072

9072

divides by

29072 = 4536 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4536

4536

divides by

24536 = 2268 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2268

2268

divides by

22268 = 1134 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 1134

1134

divides by

21134 = 567 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 567

567

divides by

3567 = 189 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 189

189

divides by

3189 = 63 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 63

63

divides by

363 = 21 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 21

21

divides by

321 = 7 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 7

2, 3, 7

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 7

= 2^{6} \cdot 3^{4} \cdot 7

= 2^{6} \cdot 3^{4} \cdot 7

Apply radical rule:

= 7 2^{6} 3^{4}

Apply radical rule:

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

= 2^{3} 7 3^{4}

Apply radical rule:

3^{4} = 3^{\frac{4}{2}} = 3^{2}

= 2^{3} \cdot 3^{2} 7

Refine

= 727

u_{1, 2} = \frac{- ( - 90 ) \pm 72 7}{2 \cdot 261}

Separate the solutions

u_{1} = \frac{- ( - 90 ) + 72 7}{2 \cdot 261}, u_{2} = \frac{- ( - 90 ) - 72 7}{2 \cdot 261}

u = \frac{- ( - 90 ) + 72 7}{2 \cdot 261} : \frac{5 + 4 7}{29}

\frac{- ( - 90 ) + 72 7}{2 \cdot 261}

Apply rule

- (- a) = a

= \frac{90 + 72 7}{2 \cdot 261}

Multiply the numbers:

2 \cdot 261 = 522

= \frac{90 + 72 7}{522}

Factor

90 + 727 : 18 (5 + 47)

90 + 727

Rewrite as

= 18 \cdot 5 + 18 \cdot 47

Factor out common term

18

= 18 (5 + 47)

= \frac{18 ( 5 + 4 7 )}{522}

Cancel the common factor:

18

= \frac{5 + 4 7}{29}

u = \frac{- ( - 90 ) - 72 7}{2 \cdot 261} : \frac{5 - 4 7}{29}

\frac{- ( - 90 ) - 72 7}{2 \cdot 261}

Apply rule

- (- a) = a

= \frac{90 - 72 7}{2 \cdot 261}

Multiply the numbers:

2 \cdot 261 = 522

= \frac{90 - 72 7}{522}

Factor

90 - 727 : 18 (5 - 47)

90 - 727

Rewrite as

= 18 \cdot 5 - 18 \cdot 47

Factor out common term

18

= 18 (5 - 47)

= \frac{18 ( 5 - 4 7 )}{522}

Cancel the common factor:

18

= \frac{5 - 4 7}{29}

The solutions to the quadratic equation are:

u = \frac{5 + 4 7}{29}, u = \frac{5 - 4 7}{29}

Substitute back

u = sin (x)

sin (x) = \frac{5 + 4 7}{29}, sin (x) = \frac{5 - 4 7}{29}

sin (x) = \frac{5 + 4 7}{29}, sin (x) = \frac{5 - 4 7}{29}

sin (x) = \frac{5 + 4 7}{29} : x = arcsin (\frac{5 + 4 7}{29}) + 2 πn, x = π - arcsin (\frac{5 + 4 7}{29}) + 2 πn

sin (x) = \frac{5 + 4 7}{29}

Apply trig inverse properties

sin (x) = \frac{5 + 4 7}{29}

General solutions for

sin (x) = \frac{5 + 4 7}{29}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{5 + 4 7}{29}) + 2 πn, x = π - arcsin (\frac{5 + 4 7}{29}) + 2 πn

x = arcsin (\frac{5 + 4 7}{29}) + 2 πn, x = π - arcsin (\frac{5 + 4 7}{29}) + 2 πn

sin (x) = \frac{5 - 4 7}{29} : x = arcsin (\frac{5 - 4 7}{29}) + 2 πn, x = π + arcsin (- \frac{5 - 4 7}{29}) + 2 πn

sin (x) = \frac{5 - 4 7}{29}

Apply trig inverse properties

sin (x) = \frac{5 - 4 7}{29}

General solutions for

sin (x) = \frac{5 - 4 7}{29}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{5 - 4 7}{29}) + 2 πn, x = π + arcsin (- \frac{5 - 4 7}{29}) + 2 πn

x = arcsin (\frac{5 - 4 7}{29}) + 2 πn, x = π + arcsin (- \frac{5 - 4 7}{29}) + 2 πn

Combine all the solutions

x = arcsin (\frac{5 + 4 7}{29}) + 2 πn, x = π - arcsin (\frac{5 + 4 7}{29}) + 2 πn, x = arcsin (\frac{5 - 4 7}{29}) + 2 πn, x = π + arcsin (- \frac{5 - 4 7}{29}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

15 sin (x) + 6 cos (x) - 3 = 0

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

Check the solution

arcsin (\frac{5 + 4 7}{29}) + 2 πn :

False

arcsin (\frac{5 + 4 7}{29}) + 2 πn

Plug in

n = 1

arcsin (\frac{5 + 4 7}{29}) + 2 π 1

For

15 sin (x) + 6 cos (x) - 3 = 0

plug in

x = arcsin (\frac{5 + 4 7}{29}) + 2 π 1

15 sin (arcsin (\frac{5 + 4 7}{29}) + 2 π 1) + 6 cos (arcsin (\frac{5 + 4 7}{29}) + 2 π 1) - 3 = 0

Refine

10.12035 \dots = 0

\Rightarrow False

Check the solution

π - arcsin (\frac{5 + 4 7}{29}) + 2 πn :

True

π - arcsin (\frac{5 + 4 7}{29}) + 2 πn

Plug in

n = 1

π - arcsin (\frac{5 + 4 7}{29}) + 2 π 1

For

15 sin (x) + 6 cos (x) - 3 = 0

plug in

x = π - arcsin (\frac{5 + 4 7}{29}) + 2 π 1

15 sin (π - arcsin (\frac{5 + 4 7}{29}) + 2 π 1) + 6 cos (π - arcsin (\frac{5 + 4 7}{29}) + 2 π 1) - 3 = 0

Refine

0 = 0

\Rightarrow True

Check the solution

arcsin (\frac{5 - 4 7}{29}) + 2 πn :

True

arcsin (\frac{5 - 4 7}{29}) + 2 πn

Plug in

n = 1

arcsin (\frac{5 - 4 7}{29}) + 2 π 1

For

15 sin (x) + 6 cos (x) - 3 = 0

plug in

x = arcsin (\frac{5 - 4 7}{29}) + 2 π 1

15 sin (arcsin (\frac{5 - 4 7}{29}) + 2 π 1) + 6 cos (arcsin (\frac{5 - 4 7}{29}) + 2 π 1) - 3 = 0

Refine

0 = 0

\Rightarrow True

Check the solution

π + arcsin (- \frac{5 - 4 7}{29}) + 2 πn :

False

π + arcsin (- \frac{5 - 4 7}{29}) + 2 πn

Plug in

n = 1

π + arcsin (- \frac{5 - 4 7}{29}) + 2 π 1

For

15 sin (x) + 6 cos (x) - 3 = 0

plug in

x = π + arcsin (- \frac{5 - 4 7}{29}) + 2 π 1

15 sin (π + arcsin (- \frac{5 - 4 7}{29}) + 2 π 1) + 6 cos (π + arcsin (- \frac{5 - 4 7}{29}) + 2 π 1) - 3 = 0

Refine

- 11.77552 \dots = 0

\Rightarrow False

x = π - arcsin (\frac{5 + 4 7}{29}) + 2 πn, x = arcsin (\frac{5 - 4 7}{29}) + 2 πn

Show solutions in decimal form

x = π - 0.56728 \dots + 2 πn, x = - 0.19372 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 15sin(x)+6cos(x)-3=0 ?
The general solution for 15sin(x)+6cos(x)-3=0 is x=pi-0.56728…+2pin,x=-0.19372…+2pin