What is the general solution for 3cos(x)-2sin(x)=1 ?

The general solution for 3cos(x)-2sin(x)=1 is x=pi+1.26382…+2pin,x=0.70175…+2pin

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Popular >

3cos(x)-2sin(x)=1

Solution

3 cos (x) - 2 sin (x) = 1

Solution

x = π + 1.26382 \dots + 2 πn, x = 0.70175 \dots + 2 πn

Degrees

x = 252.41204 \dots^{\circ} + 36 0^{\circ} n, x = 40.20781 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 cos (x) - 2 sin (x) = 1

Add

2 sin (x)

to both sides

3 cos (x) = 1 + 2 sin (x)

Square both sides

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

Subtract

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

from both sides

9 cos^{2} (x) - 1 - 4 sin (x) - 4 sin^{2} (x) = 0

Rewrite using trig identities

- 1 - 4 sin (x) - 4 sin^{2} (x) + 9 cos^{2} (x)

Use the Pythagorean identity:

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

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

= - 1 - 4 sin (x) - 4 sin^{2} (x) + 9 (1 - sin^{2} (x))

Simplify

- 1 - 4 sin (x) - 4 sin^{2} (x) + 9 (1 - sin^{2} (x)) : - 13 sin^{2} (x) - 4 sin (x) + 8

- 1 - 4 sin (x) - 4 sin^{2} (x) + 9 (1 - sin^{2} (x))

Expand

9 (1 - sin^{2} (x)) : 9 - 9 sin^{2} (x)

9 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 9, b = 1, c = sin^{2} (x)

= 9 \cdot 1 - 9 sin^{2} (x)

Multiply the numbers:

9 \cdot 1 = 9

= 9 - 9 sin^{2} (x)

= - 1 - 4 sin (x) - 4 sin^{2} (x) + 9 - 9 sin^{2} (x)

Simplify

- 1 - 4 sin (x) - 4 sin^{2} (x) + 9 - 9 sin^{2} (x) : - 13 sin^{2} (x) - 4 sin (x) + 8

- 1 - 4 sin (x) - 4 sin^{2} (x) + 9 - 9 sin^{2} (x)

Group like terms

= - 4 sin (x) - 4 sin^{2} (x) - 9 sin^{2} (x) - 1 + 9

Add similar elements:

- 4 sin^{2} (x) - 9 sin^{2} (x) = - 13 sin^{2} (x)

= - 4 sin (x) - 13 sin^{2} (x) - 1 + 9

Add/Subtract the numbers:

- 1 + 9 = 8

= - 13 sin^{2} (x) - 4 sin (x) + 8

= - 13 sin^{2} (x) - 4 sin (x) + 8

= - 13 sin^{2} (x) - 4 sin (x) + 8

8 - 13 sin^{2} (x) - 4 sin (x) = 0

Solve by substitution

8 - 13 sin^{2} (x) - 4 sin (x) = 0

Let:

sin (x) = u

8 - 13 u^{2} - 4 u = 0

8 - 13 u^{2} - 4 u = 0 : u = - \frac{2 ( 1 + 3 3 )}{13}, u = \frac{2 ( 3 3 - 1 )}{13}

8 - 13 u^{2} - 4 u = 0

Write in the standard form

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

- 13 u^{2} - 4 u + 8 = 0

Solve with the quadratic formula

- 13 u^{2} - 4 u + 8 = 0

Quadratic Equation Formula:

For

a = - 13, b = - 4, c = 8

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 13 ) \cdot 8}{2 ( - 13 )}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 13 ) \cdot 8}{2 ( - 13 )}

(- 4)^{2} - 4 (- 13) \cdot 8 = 123

(- 4)^{2} - 4 (- 13) \cdot 8

Apply rule

- (- a) = a

= (- 4)^{2} + 4 \cdot 13 \cdot 8

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

= 4^{2} + 4 \cdot 13 \cdot 8

Multiply the numbers:

4 \cdot 13 \cdot 8 = 416

= 4^{2} + 416

4^{2} = 16

= 16 + 416

Add the numbers:

16 + 416 = 432

= 432

Prime factorization of

432 : 2^{4} \cdot 3^{3}

432

432

divides by

2432 = 216 \cdot 2

= 2 \cdot 216

216

divides by

2216 = 108 \cdot 2

= 2 \cdot 2 \cdot 108

108

divides by

2108 = 54 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 54

54

divides by

254 = 27 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 27

27

divides by

327 = 9 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 9

9

divides by

39 = 3 \cdot 3

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

2, 3

are all prime numbers, therefore no further factorization is possible

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

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

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

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

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

Apply radical rule:

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

Apply radical rule:

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

= 2^{2} 3 3^{2}

Apply radical rule:

3^{2} = 3

= 2^{2} \cdot 33

Refine

= 123

u_{1, 2} = \frac{- ( - 4 ) \pm 12 3}{2 ( - 13 )}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 12 3}{2 ( - 13 )}, u_{2} = \frac{- ( - 4 ) - 12 3}{2 ( - 13 )}

u = \frac{- ( - 4 ) + 12 3}{2 ( - 13 )} : - \frac{2 ( 1 + 3 3 )}{13}

\frac{- ( - 4 ) + 12 3}{2 ( - 13 )}

Remove parentheses:

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

= \frac{4 + 12 3}{- 2 \cdot 13}

Multiply the numbers:

2 \cdot 13 = 26

= \frac{4 + 12 3}{- 26}

Apply the fraction rule:

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

= - \frac{4 + 12 3}{26}

Cancel

\frac{4 + 12 3}{26} : \frac{2 ( 1 + 3 3 )}{13}

\frac{4 + 12 3}{26}

Factor

4 + 123 : 4 (1 + 33)

4 + 123

Rewrite as

= 4 \cdot 1 + 4 \cdot 33

Factor out common term

4

= 4 (1 + 33)

= \frac{4 ( 1 + 3 3 )}{26}

Cancel the common factor:

2

= \frac{2 ( 1 + 3 3 )}{13}

= - \frac{2 ( 1 + 3 3 )}{13}

u = \frac{- ( - 4 ) - 12 3}{2 ( - 13 )} : \frac{2 ( 3 3 - 1 )}{13}

\frac{- ( - 4 ) - 12 3}{2 ( - 13 )}

Remove parentheses:

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

= \frac{4 - 12 3}{- 2 \cdot 13}

Multiply the numbers:

2 \cdot 13 = 26

= \frac{4 - 12 3}{- 26}

Apply the fraction rule:

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

4 - 123 = - (123 - 4)

= \frac{12 3 - 4}{26}

Factor

123 - 4 : 4 (33 - 1)

123 - 4

Rewrite as

= 4 \cdot 33 - 4 \cdot 1

Factor out common term

4

= 4 (33 - 1)

= \frac{4 ( 3 3 - 1 )}{26}

Cancel the common factor:

2

= \frac{2 ( 3 3 - 1 )}{13}

The solutions to the quadratic equation are:

u = - \frac{2 ( 1 + 3 3 )}{13}, u = \frac{2 ( 3 3 - 1 )}{13}

Substitute back

u = sin (x)

sin (x) = - \frac{2 ( 1 + 3 3 )}{13}, sin (x) = \frac{2 ( 3 3 - 1 )}{13}

sin (x) = - \frac{2 ( 1 + 3 3 )}{13}, sin (x) = \frac{2 ( 3 3 - 1 )}{13}

sin (x) = - \frac{2 ( 1 + 3 3 )}{13} : x = arcsin (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 πn

sin (x) = - \frac{2 ( 1 + 3 3 )}{13}

Apply trig inverse properties

sin (x) = - \frac{2 ( 1 + 3 3 )}{13}

General solutions for

sin (x) = - \frac{2 ( 1 + 3 3 )}{13}

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

x = arcsin (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 πn

x = arcsin (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 πn

sin (x) = \frac{2 ( 3 3 - 1 )}{13} : x = arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn, x = π - arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

sin (x) = \frac{2 ( 3 3 - 1 )}{13}

Apply trig inverse properties

sin (x) = \frac{2 ( 3 3 - 1 )}{13}

General solutions for

sin (x) = \frac{2 ( 3 3 - 1 )}{13}

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

x = arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn, x = π - arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

x = arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn, x = π - arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

Combine all the solutions

x = arcsin (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn, x = π - arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

3 cos (x) - 2 sin (x) = 1

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

Check the solution

arcsin (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn :

False

arcsin (- \frac{2 ( 1 + 3 3 )}{13}) + 2 πn

Plug in

n = 1

arcsin (- \frac{2 ( 1 + 3 3 )}{13}) + 2 π 1

For

3 cos (x) - 2 sin (x) = 1

plug in

x = arcsin (- \frac{2 ( 1 + 3 3 )}{13}) + 2 π 1

3 cos (arcsin (- \frac{2 ( 1 + 3 3 )}{13}) + 2 π 1) - 2 sin (arcsin (- \frac{2 ( 1 + 3 3 )}{13}) + 2 π 1) = 1

Refine

2.81301 \dots = 1

\Rightarrow False

Check the solution

π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 πn :

True

π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 πn

Plug in

n = 1

π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 π 1

For

3 cos (x) - 2 sin (x) = 1

plug in

x = π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 π 1

3 cos (π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 π 1) - 2 sin (π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 π 1) = 1

Refine

1 = 1

\Rightarrow True

Check the solution

arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn :

True

arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

Plug in

n = 1

arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1

For

3 cos (x) - 2 sin (x) = 1

plug in

x = arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1

3 cos (arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1) - 2 sin (arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1) = 1

Refine

1 = 1

\Rightarrow True

Check the solution

π - arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn :

False

π - arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

Plug in

n = 1

π - arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1

For

3 cos (x) - 2 sin (x) = 1

plug in

x = π - arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1

3 cos (π - arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1) - 2 sin (π - arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 π 1) = 1

Refine

- 3.58224 \dots = 1

\Rightarrow False

x = π + arcsin (\frac{2 ( 1 + 3 3 )}{13}) + 2 πn, x = arcsin (\frac{2 ( 3 3 - 1 )}{13}) + 2 πn

Show solutions in decimal form

x = π + 1.26382 \dots + 2 πn, x = 0.70175 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos(x)-2sin(x)=1 ?
The general solution for 3cos(x)-2sin(x)=1 is x=pi+1.26382…+2pin,x=0.70175…+2pin