What is the general solution for 2sin(x)+5cos(x)=4 ?

The general solution for 2sin(x)+5cos(x)=4 is x=1.11408…+2pin,x=2pi-0.35307…+2pin

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Popular Trigonometry >

2sin(x)+5cos(x)=4

Solution

2 sin (x) + 5 cos (x) = 4

Solution

x = 1.11408 \dots + 2 πn, x = 2 π - 0.35307 \dots + 2 πn

Degrees

x = 63.83252 \dots^{\circ} + 36 0^{\circ} n, x = 339.77029 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (x) + 5 cos (x) = 4

Subtract

5 cos (x)

from both sides

2 sin (x) = 4 - 5 cos (x)

Square both sides

(2 sin (x))^{2} = (4 - 5 cos (x))^{2}

Subtract

(4 - 5 cos (x))^{2}

from both sides

4 sin^{2} (x) - 16 + 40 cos (x) - 25 cos^{2} (x) = 0

Rewrite using trig identities

- 16 - 25 cos^{2} (x) + 40 cos (x) + 4 sin^{2} (x)

Use the Pythagorean identity:

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

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

= - 16 - 25 cos^{2} (x) + 40 cos (x) + 4 (1 - cos^{2} (x))

Simplify

- 16 - 25 cos^{2} (x) + 40 cos (x) + 4 (1 - cos^{2} (x)) : 40 cos (x) - 29 cos^{2} (x) - 12

- 16 - 25 cos^{2} (x) + 40 cos (x) + 4 (1 - cos^{2} (x))

Expand

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

4 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

= 4 \cdot 1 - 4 cos^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

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

= - 16 - 25 cos^{2} (x) + 40 cos (x) + 4 - 4 cos^{2} (x)

Simplify

- 16 - 25 cos^{2} (x) + 40 cos (x) + 4 - 4 cos^{2} (x) : 40 cos (x) - 29 cos^{2} (x) - 12

- 16 - 25 cos^{2} (x) + 40 cos (x) + 4 - 4 cos^{2} (x)

Group like terms

= - 25 cos^{2} (x) + 40 cos (x) - 4 cos^{2} (x) - 16 + 4

Add similar elements:

- 25 cos^{2} (x) - 4 cos^{2} (x) = - 29 cos^{2} (x)

= - 29 cos^{2} (x) + 40 cos (x) - 16 + 4

Add/Subtract the numbers:

- 16 + 4 = - 12

= 40 cos (x) - 29 cos^{2} (x) - 12

= 40 cos (x) - 29 cos^{2} (x) - 12

= 40 cos (x) - 29 cos^{2} (x) - 12

- 12 - 29 cos^{2} (x) + 40 cos (x) = 0

Solve by substitution

- 12 - 29 cos^{2} (x) + 40 cos (x) = 0

Let:

cos (x) = u

- 12 - 29 u^{2} + 40 u = 0

- 12 - 29 u^{2} + 40 u = 0 : u = \frac{2 ( 10 - 13 )}{29}, u = \frac{2 ( 10 + 13 )}{29}

- 12 - 29 u^{2} + 40 u = 0

Write in the standard form

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

- 29 u^{2} + 40 u - 12 = 0

Solve with the quadratic formula

- 29 u^{2} + 40 u - 12 = 0

Quadratic Equation Formula:

For

a = - 29, b = 40, c = - 12

u_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 29 ) ( - 12 )}{2 ( - 29 )}

u_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 29 ) ( - 12 )}{2 ( - 29 )}

4 0^{2} - 4 (- 29) (- 12) = 413

4 0^{2} - 4 (- 29) (- 12)

Apply rule

- (- a) = a

= 4 0^{2} - 4 \cdot 29 \cdot 12

Multiply the numbers:

4 \cdot 29 \cdot 12 = 1392

= 4 0^{2} - 1392

4 0^{2} = 1600

= 1600 - 1392

Subtract the numbers:

1600 - 1392 = 208

= 208

Prime factorization of

208 : 2^{4} \cdot 13

208

208

divides by

2208 = 104 \cdot 2

= 2 \cdot 104

104

divides by

2104 = 52 \cdot 2

= 2 \cdot 2 \cdot 52

52

divides by

252 = 26 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 26

26

divides by

226 = 13 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 13

2, 13

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 13

= 2^{4} \cdot 13

= 2^{4} \cdot 13

Apply radical rule:

n ab = n a n b

= 13 2^{4}

Apply radical rule:

n a^{m} = a^{\frac{m}{n}}

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

= 2^{2} 13

Refine

= 413

u_{1, 2} = \frac{- 40 \pm 4 13}{2 ( - 29 )}

Separate the solutions

u_{1} = \frac{- 40 + 4 13}{2 ( - 29 )}, u_{2} = \frac{- 40 - 4 13}{2 ( - 29 )}

u = \frac{- 40 + 4 13}{2 ( - 29 )} : \frac{2 ( 10 - 13 )}{29}

\frac{- 40 + 4 13}{2 ( - 29 )}

Remove parentheses:

(- a) = - a

= \frac{- 40 + 4 13}{- 2 \cdot 29}

Multiply the numbers:

2 \cdot 29 = 58

= \frac{- 40 + 4 13}{- 58}

Apply the fraction rule:

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

- 40 + 413 = - (40 - 413)

= \frac{40 - 4 13}{58}

Factor

40 - 413 : 4 (10 - 13)

40 - 413

Rewrite as

= 4 \cdot 10 - 413

Factor out common term

4

= 4 (10 - 13)

= \frac{4 ( 10 - 13 )}{58}

Cancel the common factor:

2

= \frac{2 ( 10 - 13 )}{29}

u = \frac{- 40 - 4 13}{2 ( - 29 )} : \frac{2 ( 10 + 13 )}{29}

\frac{- 40 - 4 13}{2 ( - 29 )}

Remove parentheses:

(- a) = - a

= \frac{- 40 - 4 13}{- 2 \cdot 29}

Multiply the numbers:

2 \cdot 29 = 58

= \frac{- 40 - 4 13}{- 58}

Apply the fraction rule:

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

- 40 - 413 = - (40 + 413)

= \frac{40 + 4 13}{58}

Factor

40 + 413 : 4 (10 + 13)

40 + 413

Rewrite as

= 4 \cdot 10 + 413

Factor out common term

4

= 4 (10 + 13)

= \frac{4 ( 10 + 13 )}{58}

Cancel the common factor:

2

= \frac{2 ( 10 + 13 )}{29}

The solutions to the quadratic equation are:

u = \frac{2 ( 10 - 13 )}{29}, u = \frac{2 ( 10 + 13 )}{29}

Substitute back

u = cos (x)

cos (x) = \frac{2 ( 10 - 13 )}{29}, cos (x) = \frac{2 ( 10 + 13 )}{29}

cos (x) = \frac{2 ( 10 - 13 )}{29}, cos (x) = \frac{2 ( 10 + 13 )}{29}

cos (x) = \frac{2 ( 10 - 13 )}{29} : x = arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn, x = 2 π - arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn

cos (x) = \frac{2 ( 10 - 13 )}{29}

Apply trig inverse properties

cos (x) = \frac{2 ( 10 - 13 )}{29}

General solutions for

cos (x) = \frac{2 ( 10 - 13 )}{29}

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

x = arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn, x = 2 π - arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn

x = arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn, x = 2 π - arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn

cos (x) = \frac{2 ( 10 + 13 )}{29} : x = arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn, x = 2 π - arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn

cos (x) = \frac{2 ( 10 + 13 )}{29}

Apply trig inverse properties

cos (x) = \frac{2 ( 10 + 13 )}{29}

General solutions for

cos (x) = \frac{2 ( 10 + 13 )}{29}

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

x = arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn, x = 2 π - arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn

x = arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn, x = 2 π - arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn

Combine all the solutions

x = arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn, x = 2 π - arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn, x = arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn, x = 2 π - arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

2 sin (x) + 5 cos (x) = 4

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

Check the solution

arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn :

True

arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn

Plug in

n = 1

arccos (\frac{2 ( 10 - 13 )}{29}) + 2 π 1

For

2 sin (x) + 5 cos (x) = 4

plug in

x = arccos (\frac{2 ( 10 - 13 )}{29}) + 2 π 1

2 sin (arccos (\frac{2 ( 10 - 13 )}{29}) + 2 π 1) + 5 cos (arccos (\frac{2 ( 10 - 13 )}{29}) + 2 π 1) = 4

Refine

4 = 4

\Rightarrow True

Check the solution

2 π - arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn :

False

2 π - arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{2 ( 10 - 13 )}{29}) + 2 π 1

For

2 sin (x) + 5 cos (x) = 4

plug in

x = 2 π - arccos (\frac{2 ( 10 - 13 )}{29}) + 2 π 1

2 sin (2 π - arccos (\frac{2 ( 10 - 13 )}{29}) + 2 π 1) + 5 cos (2 π - arccos (\frac{2 ( 10 - 13 )}{29}) + 2 π 1) = 4

Refine

0.40996 \dots = 4

\Rightarrow False

Check the solution

arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn :

False

arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn

Plug in

n = 1

arccos (\frac{2 ( 10 + 13 )}{29}) + 2 π 1

For

2 sin (x) + 5 cos (x) = 4

plug in

x = arccos (\frac{2 ( 10 + 13 )}{29}) + 2 π 1

2 sin (arccos (\frac{2 ( 10 + 13 )}{29}) + 2 π 1) + 5 cos (arccos (\frac{2 ( 10 + 13 )}{29}) + 2 π 1) = 4

Refine

5.38313 \dots = 4

\Rightarrow False

Check the solution

2 π - arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn :

True

2 π - arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{2 ( 10 + 13 )}{29}) + 2 π 1

For

2 sin (x) + 5 cos (x) = 4

plug in

x = 2 π - arccos (\frac{2 ( 10 + 13 )}{29}) + 2 π 1

2 sin (2 π - arccos (\frac{2 ( 10 + 13 )}{29}) + 2 π 1) + 5 cos (2 π - arccos (\frac{2 ( 10 + 13 )}{29}) + 2 π 1) = 4

Refine

4 = 4

\Rightarrow True

x = arccos (\frac{2 ( 10 - 13 )}{29}) + 2 πn, x = 2 π - arccos (\frac{2 ( 10 + 13 )}{29}) + 2 πn

Show solutions in decimal form

x = 1.11408 \dots + 2 πn, x = 2 π - 0.35307 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(x)+5cos(x)=4 ?
The general solution for 2sin(x)+5cos(x)=4 is x=1.11408…+2pin,x=2pi-0.35307…+2pin