What is the general solution for sqrt(3)sin(x)+cos(x)=4 ?

The general solution for sqrt(3)sin(x)+cos(x)=4 is No Solution for x\in\mathbb{R}

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

cos (x)

from both sides

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

Square both sides

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

Subtract

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

from both sides

3 sin^{2} (x) - 16 + 8 cos (x) - cos^{2} (x) = 0

Rewrite using trig identities

- 16 - cos^{2} (x) + 3 sin^{2} (x) + 8 cos (x)

Use the Pythagorean identity:

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

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

= - 16 - cos^{2} (x) + 3 (1 - cos^{2} (x)) + 8 cos (x)

Simplify

- 16 - cos^{2} (x) + 3 (1 - cos^{2} (x)) + 8 cos (x) : 8 cos (x) - 4 cos^{2} (x) - 13

- 16 - cos^{2} (x) + 3 (1 - cos^{2} (x)) + 8 cos (x)

Expand

3 (1 - cos^{2} (x)) : 3 - 3 cos^{2} (x)

3 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = cos^{2} (x)

= 3 \cdot 1 - 3 cos^{2} (x)

Multiply the numbers:

3 \cdot 1 = 3

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

= - 16 - cos^{2} (x) + 3 - 3 cos^{2} (x) + 8 cos (x)

Simplify

- 16 - cos^{2} (x) + 3 - 3 cos^{2} (x) + 8 cos (x) : 8 cos (x) - 4 cos^{2} (x) - 13

- 16 - cos^{2} (x) + 3 - 3 cos^{2} (x) + 8 cos (x)

Group like terms

= - cos^{2} (x) - 3 cos^{2} (x) + 8 cos (x) - 16 + 3

Add similar elements:

- cos^{2} (x) - 3 cos^{2} (x) = - 4 cos^{2} (x)

= - 4 cos^{2} (x) + 8 cos (x) - 16 + 3

Add/Subtract the numbers:

- 16 + 3 = - 13

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

Simplify

8^{2} - 4 (- 4) (- 13) : 12 i

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 13 = 208

= 8^{2} - 208

Apply imaginary number rule:

- a = i a

= i 208 - 8^{2}

- 8^{2} + 208 = 12

- 8^{2} + 208

8^{2} = 64

= - 64 + 208

Add/Subtract the numbers:

- 64 + 208 = 144

= 144

Factor the number:

144 = 1 2^{2}

= 1 2^{2}

Apply radical rule:

1 2^{2} = 12

= 12

= 12 i

u_{1, 2} = \frac{- 8 \pm 12 i}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 8 + 12 i}{2 ( - 4 )}, u_{2} = \frac{- 8 - 12 i}{2 ( - 4 )}

u = \frac{- 8 + 12 i}{2 ( - 4 )} : 1 - i \frac{3}{2}

\frac{- 8 + 12 i}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 8 + 12 i}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8 + 12 i}{- 8}

Apply the fraction rule:

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

= - \frac{- 8 + 12 i}{8}

Cancel

\frac{- 8 + 12 i}{8} : \frac{- 2 + 3 i}{2}

\frac{- 8 + 12 i}{8}

Factor

- 8 + 12 i : 4 (- 2 + 3 i)

- 8 + 12 i

Rewrite as

= - 4 \cdot 2 + 4 \cdot 3 i

Factor out common term

4

= 4 (- 2 + 3 i)

= \frac{4 ( - 2 + 3 i )}{8}

Cancel the common factor:

4

= \frac{- 2 + 3 i}{2}

= - \frac{- 2 + 3 i}{2}

Rewrite

- \frac{- 2 + 3 i}{2}

in standard complex form:

1 - \frac{3}{2} i

- \frac{- 2 + 3 i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 2 + 3 i}{2} = - (- \frac{2}{2}) - (\frac{3 i}{2})

= - (- \frac{2}{2}) - (\frac{3 i}{2})

Remove parentheses:

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

= \frac{2}{2} - \frac{3 i}{2}

Apply rule

\frac{a}{a} = 1

= 1 - \frac{3 i}{2}

= 1 - \frac{3}{2} i

u = \frac{- 8 - 12 i}{2 ( - 4 )} : 1 + i \frac{3}{2}

\frac{- 8 - 12 i}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 8 - 12 i}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 8 - 12 i}{- 8}

Apply the fraction rule:

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

= - \frac{- 8 - 12 i}{8}

Cancel

\frac{- 8 - 12 i}{8} : - \frac{2 + 3 i}{2}

\frac{- 8 - 12 i}{8}

Factor

- 8 - 12 i : - 4 (2 + 3 i)

- 8 - 12 i

Rewrite as

= - 4 \cdot 2 - 4 \cdot 3 i

Factor out common term

4

= - 4 (2 + 3 i)

= - \frac{4 ( 2 + 3 i )}{8}

Cancel the common factor:

4

= - \frac{2 + 3 i}{2}

= - (- \frac{2 + 3 i}{2})

Apply rule

- (- a) = a

= \frac{2 + 3 i}{2}

Rewrite

\frac{2 + 3 i}{2}

in standard complex form:

1 + \frac{3}{2} i

\frac{2 + 3 i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{2 + 3 i}{2} = \frac{2}{2} + \frac{3 i}{2}

= \frac{2}{2} + \frac{3 i}{2}

Apply rule

\frac{a}{a} = 1

\frac{2}{2} = 1

= 1 + \frac{3 i}{2}

= 1 + \frac{3}{2} i

The solutions to the quadratic equation are:

u = 1 - i \frac{3}{2}, u = 1 + i \frac{3}{2}

Substitute back

u = cos (x)

cos (x) = 1 - i \frac{3}{2}, cos (x) = 1 + i \frac{3}{2}

cos (x) = 1 - i \frac{3}{2}, cos (x) = 1 + i \frac{3}{2}

cos (x) = 1 - i \frac{3}{2} :

No Solution

cos (x) = 1 - i \frac{3}{2}

No Solution

cos (x) = 1 + i \frac{3}{2} :

No Solution

cos (x) = 1 + i \frac{3}{2}

No Solution

Combine all the solutions

No Solution

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)sin(x)+cos(x)=4 ?
The general solution for sqrt(3)sin(x)+cos(x)=4 is No Solution for x\in\mathbb{R}