What is the general solution for 2sin(x)-4cos(x)=3,0<= x<= 2pi ?

The general solution for 2sin(x)-4cos(x)=3,0<= x<= 2pi is x=-2.76975…+2pi,x=1.84246…

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2sin(x)-4cos(x)=3,0<= x<= 2pi

Solution

2 sin (x) - 4 cos (x) = 3, 0 \leq x \leq 2 π

Solution

x = - 2.76975 \dots + 2 π, x = 1.84246 \dots

Degrees

x = 201.30453 \dots^{\circ}, x = 105.56536 \dots^{\circ}

Solution steps

2 sin (x) - 4 cos (x) = 3, 0 \leq x \leq 2 π

Add

4 cos (x)

to both sides

2 sin (x) = 3 + 4 cos (x)

Square both sides

(2 sin (x))^{2} = (3 + 4 cos (x))^{2}

Subtract

(3 + 4 cos (x))^{2}

from both sides

4 sin^{2} (x) - 9 - 24 cos (x) - 16 cos^{2} (x) = 0

Rewrite using trig identities

- 9 - 16 cos^{2} (x) - 24 cos (x) + 4 sin^{2} (x)

Use the Pythagorean identity:

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

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

= - 9 - 16 cos^{2} (x) - 24 cos (x) + 4 (1 - cos^{2} (x))

Simplify

- 9 - 16 cos^{2} (x) - 24 cos (x) + 4 (1 - cos^{2} (x)) : - 20 cos^{2} (x) - 24 cos (x) - 5

- 9 - 16 cos^{2} (x) - 24 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)

= - 9 - 16 cos^{2} (x) - 24 cos (x) + 4 - 4 cos^{2} (x)

Simplify

- 9 - 16 cos^{2} (x) - 24 cos (x) + 4 - 4 cos^{2} (x) : - 20 cos^{2} (x) - 24 cos (x) - 5

- 9 - 16 cos^{2} (x) - 24 cos (x) + 4 - 4 cos^{2} (x)

Group like terms

= - 16 cos^{2} (x) - 24 cos (x) - 4 cos^{2} (x) - 9 + 4

Add similar elements:

- 16 cos^{2} (x) - 4 cos^{2} (x) = - 20 cos^{2} (x)

= - 20 cos^{2} (x) - 24 cos (x) - 9 + 4

Add/Subtract the numbers:

- 9 + 4 = - 5

= - 20 cos^{2} (x) - 24 cos (x) - 5

= - 20 cos^{2} (x) - 24 cos (x) - 5

= - 20 cos^{2} (x) - 24 cos (x) - 5

- 5 - 20 cos^{2} (x) - 24 cos (x) = 0

Solve by substitution

- 5 - 20 cos^{2} (x) - 24 cos (x) = 0

Let:

cos (x) = u

- 5 - 20 u^{2} - 24 u = 0

- 5 - 20 u^{2} - 24 u = 0 : u = - \frac{6 + 11}{10}, u = - \frac{6 - 11}{10}

- 5 - 20 u^{2} - 24 u = 0

Write in the standard form

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

- 20 u^{2} - 24 u - 5 = 0

Solve with the quadratic formula

- 20 u^{2} - 24 u - 5 = 0

Quadratic Equation Formula:

For

a = - 20, b = - 24, c = - 5

u_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 ( - 20 ) ( - 5 )}{2 ( - 20 )}

u_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 ( - 20 ) ( - 5 )}{2 ( - 20 )}

(- 24)^{2} - 4 (- 20) (- 5) = 411

(- 24)^{2} - 4 (- 20) (- 5)

Apply rule

- (- a) = a

= (- 24)^{2} - 4 \cdot 20 \cdot 5

Apply exponent rule:

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

n

is even

(- 24)^{2} = 2 4^{2}

= 2 4^{2} - 4 \cdot 20 \cdot 5

Multiply the numbers:

4 \cdot 20 \cdot 5 = 400

= 2 4^{2} - 400

2 4^{2} = 576

= 576 - 400

Subtract the numbers:

576 - 400 = 176

= 176

Prime factorization of

176 : 2^{4} \cdot 11

176

176

divides by

2176 = 88 \cdot 2

= 2 \cdot 88

88

divides by

288 = 44 \cdot 2

= 2 \cdot 2 \cdot 44

44

divides by

244 = 22 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 22

22

divides by

222 = 11 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 11

2, 11

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 11

= 2^{4} \cdot 11

= 2^{4} \cdot 11

Apply radical rule:

= 11 2^{4}

Apply radical rule:

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

= 2^{2} 11

Refine

= 411

u_{1, 2} = \frac{- ( - 24 ) \pm 4 11}{2 ( - 20 )}

Separate the solutions

u_{1} = \frac{- ( - 24 ) + 4 11}{2 ( - 20 )}, u_{2} = \frac{- ( - 24 ) - 4 11}{2 ( - 20 )}

u = \frac{- ( - 24 ) + 4 11}{2 ( - 20 )} : - \frac{6 + 11}{10}

\frac{- ( - 24 ) + 4 11}{2 ( - 20 )}

Remove parentheses:

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

= \frac{24 + 4 11}{- 2 \cdot 20}

Multiply the numbers:

2 \cdot 20 = 40

= \frac{24 + 4 11}{- 40}

Apply the fraction rule:

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

= - \frac{24 + 4 11}{40}

Cancel

\frac{24 + 4 11}{40} : \frac{6 + 11}{10}

\frac{24 + 4 11}{40}

Factor

24 + 411 : 4 (6 + 11)

24 + 411

Rewrite as

= 4 \cdot 6 + 411

Factor out common term

4

= 4 (6 + 11)

= \frac{4 ( 6 + 11 )}{40}

Cancel the common factor:

4

= \frac{6 + 11}{10}

= - \frac{6 + 11}{10}

u = \frac{- ( - 24 ) - 4 11}{2 ( - 20 )} : - \frac{6 - 11}{10}

\frac{- ( - 24 ) - 4 11}{2 ( - 20 )}

Remove parentheses:

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

= \frac{24 - 4 11}{- 2 \cdot 20}

Multiply the numbers:

2 \cdot 20 = 40

= \frac{24 - 4 11}{- 40}

Apply the fraction rule:

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

= - \frac{24 - 4 11}{40}

Cancel

\frac{24 - 4 11}{40} : \frac{6 - 11}{10}

\frac{24 - 4 11}{40}

Factor

24 - 411 : 4 (6 - 11)

24 - 411

Rewrite as

= 4 \cdot 6 - 411

Factor out common term

4

= 4 (6 - 11)

= \frac{4 ( 6 - 11 )}{40}

Cancel the common factor:

4

= \frac{6 - 11}{10}

= - \frac{6 - 11}{10}

The solutions to the quadratic equation are:

u = - \frac{6 + 11}{10}, u = - \frac{6 - 11}{10}

Substitute back

u = cos (x)

cos (x) = - \frac{6 + 11}{10}, cos (x) = - \frac{6 - 11}{10}

cos (x) = - \frac{6 + 11}{10}, cos (x) = - \frac{6 - 11}{10}

cos (x) = - \frac{6 + 11}{10}, 0 \leq x \leq 2 π : x = arccos (- \frac{6 + 11}{10}), x = - arccos (- \frac{6 + 11}{10}) + 2 π

cos (x) = - \frac{6 + 11}{10}, 0 \leq x \leq 2 π

Apply trig inverse properties

cos (x) = - \frac{6 + 11}{10}

General solutions for

cos (x) = - \frac{6 + 11}{10}

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

x = arccos (- \frac{6 + 11}{10}) + 2 πn, x = - arccos (- \frac{6 + 11}{10}) + 2 πn

x = arccos (- \frac{6 + 11}{10}) + 2 πn, x = - arccos (- \frac{6 + 11}{10}) + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = arccos (- \frac{6 + 11}{10}), x = - arccos (- \frac{6 + 11}{10}) + 2 π

cos (x) = - \frac{6 - 11}{10}, 0 \leq x \leq 2 π : x = arccos (- \frac{6 - 11}{10}), x = - arccos (- \frac{6 - 11}{10}) + 2 π

cos (x) = - \frac{6 - 11}{10}, 0 \leq x \leq 2 π

Apply trig inverse properties

cos (x) = - \frac{6 - 11}{10}

General solutions for

cos (x) = - \frac{6 - 11}{10}

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

x = arccos (- \frac{6 - 11}{10}) + 2 πn, x = - arccos (- \frac{6 - 11}{10}) + 2 πn

x = arccos (- \frac{6 - 11}{10}) + 2 πn, x = - arccos (- \frac{6 - 11}{10}) + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = arccos (- \frac{6 - 11}{10}), x = - arccos (- \frac{6 - 11}{10}) + 2 π

Combine all the solutions

x = arccos (- \frac{6 + 11}{10}), x = - arccos (- \frac{6 + 11}{10}) + 2 π, x = arccos (- \frac{6 - 11}{10}), x = - arccos (- \frac{6 - 11}{10}) + 2 π

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

2 sin (x) - 4 cos (x) = 3

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

Check the solution

arccos (- \frac{6 + 11}{10}) :

False

arccos (- \frac{6 + 11}{10})

Plug in

n = 1

arccos (- \frac{6 + 11}{10})

For

2 sin (x) - 4 cos (x) = 3

plug in

x = arccos (- \frac{6 + 11}{10})

2 sin (arccos (- \frac{6 + 11}{10})) - 4 cos (arccos (- \frac{6 + 11}{10})) = 3

Refine

4.45329 \dots = 3

\Rightarrow False

Check the solution

- arccos (- \frac{6 + 11}{10}) + 2 π :

True

- arccos (- \frac{6 + 11}{10}) + 2 π

Plug in

n = 1

- arccos (- \frac{6 + 11}{10}) + 2 π

For

2 sin (x) - 4 cos (x) = 3

plug in

x = - arccos (- \frac{6 + 11}{10}) + 2 π

2 sin (- arccos (- \frac{6 + 11}{10}) + 2 π) - 4 cos (- arccos (- \frac{6 + 11}{10}) + 2 π) = 3

Refine

3 = 3

\Rightarrow True

Check the solution

arccos (- \frac{6 - 11}{10}) :

True

arccos (- \frac{6 - 11}{10})

Plug in

n = 1

arccos (- \frac{6 - 11}{10})

For

2 sin (x) - 4 cos (x) = 3

plug in

x = arccos (- \frac{6 - 11}{10})

2 sin (arccos (- \frac{6 - 11}{10})) - 4 cos (arccos (- \frac{6 - 11}{10})) = 3

Refine

3 = 3

\Rightarrow True

Check the solution

- arccos (- \frac{6 - 11}{10}) + 2 π :

False

- arccos (- \frac{6 - 11}{10}) + 2 π

Plug in

n = 1

- arccos (- \frac{6 - 11}{10}) + 2 π

For

2 sin (x) - 4 cos (x) = 3

plug in

x = - arccos (- \frac{6 - 11}{10}) + 2 π

2 sin (- arccos (- \frac{6 - 11}{10}) + 2 π) - 4 cos (- arccos (- \frac{6 - 11}{10}) + 2 π) = 3

Refine

- 0.85329 \dots = 3

\Rightarrow False

x = - arccos (- \frac{6 + 11}{10}) + 2 π, x = arccos (- \frac{6 - 11}{10})

Show solutions in decimal form

x = - 2.76975 \dots + 2 π, x = 1.84246 \dots

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(x)-4cos(x)=3,0<= x<= 2pi ?
The general solution for 2sin(x)-4cos(x)=3,0<= x<= 2pi is x=-2.76975…+2pi,x=1.84246…