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

The general solution for sin(x)+sqrt(3)cos(x)=-sqrt(2) is x=2.87979…+2pin,x=-1.83259…+2pin

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sin(x)+sqrt(3)cos(x)=-sqrt(2)

Solution

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

Solution

x = 2.87979 \dots + 2 πn, x = - 1.83259 \dots + 2 πn

Degrees

x = 16 5^{\circ} + 36 0^{\circ} n, x = - 10 5^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

3 cos (x)

from both sides

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

Square both sides

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

Subtract

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

from both sides

sin^{2} (x) - 2 - 26 cos (x) - 3 cos^{2} (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= - 2 + 1 - cos^{2} (x) - 3 cos^{2} (x) - 2 cos (x) 6

Simplify

- 2 + 1 - cos^{2} (x) - 3 cos^{2} (x) - 2 cos (x) 6 : - 4 cos^{2} (x) - 26 cos (x) - 1

- 2 + 1 - cos^{2} (x) - 3 cos^{2} (x) - 2 cos (x) 6

Add similar elements:

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

= - 2 + 1 - 4 cos^{2} (x) - 26 cos (x)

Add/Subtract the numbers:

- 2 + 1 = - 1

= - 4 cos^{2} (x) - 26 cos (x) - 1

= - 4 cos^{2} (x) - 26 cos (x) - 1

- 1 - 4 cos^{2} (x) - 2 cos (x) 6 = 0

Solve by substitution

- 1 - 4 cos^{2} (x) - 2 cos (x) 6 = 0

Let:

cos (x) = u

- 1 - 4 u^{2} - 2 u 6 = 0

- 1 - 4 u^{2} - 2 u 6 = 0 : u = - \frac{6 + 2}{4}, u = - \frac{6 - 2}{4}

- 1 - 4 u^{2} - 2 u 6 = 0

Write in the standard form

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

- 4 u^{2} - 26 u - 1 = 0

Solve with the quadratic formula

- 4 u^{2} - 26 u - 1 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 26, c = - 1

u_{1, 2} = \frac{- ( - 2 6 ) \pm ( - 2 6 ) ^{2} - 4 ( - 4 ) ( - 1 )}{2 ( - 4 )}

u_{1, 2} = \frac{- ( - 2 6 ) \pm ( - 2 6 ) ^{2} - 4 ( - 4 ) ( - 1 )}{2 ( - 4 )}

(- 26)^{2} - 4 (- 4) (- 1) = 22

(- 26)^{2} - 4 (- 4) (- 1)

Apply rule

- (- a) = a

= (- 26)^{2} - 4 \cdot 4 \cdot 1

(- 26)^{2} = 2^{2} \cdot 6

(- 26)^{2}

Apply exponent rule:

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

n

is even

(- 26)^{2} = (26)^{2}

= (26)^{2}

Apply exponent rule:

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

= 2^{2} (6)^{2}

(6)^{2} : 6

Apply radical rule:

a = a^{\frac{1}{2}}

= (6^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 6^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 6

= 2^{2} \cdot 6

4 \cdot 4 \cdot 1 = 16

4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 16

= 2^{2} \cdot 6 - 16

2^{2} \cdot 6 = 24

2^{2} \cdot 6

2^{2} = 4

= 4 \cdot 6

Multiply the numbers:

4 \cdot 6 = 24

= 24

= 24 - 16

Subtract the numbers:

24 - 16 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

u_{1, 2} = \frac{- ( - 2 6 ) \pm 2 2}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 2 6 ) + 2 2}{2 ( - 4 )}, u_{2} = \frac{- ( - 2 6 ) - 2 2}{2 ( - 4 )}

u = \frac{- ( - 2 6 ) + 2 2}{2 ( - 4 )} : - \frac{6 + 2}{4}

\frac{- ( - 2 6 ) + 2 2}{2 ( - 4 )}

Remove parentheses:

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

= \frac{2 6 + 2 2}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 6 + 2 2}{- 8}

Apply the fraction rule:

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

= - \frac{2 6 + 2 2}{8}

Cancel

\frac{2 6 + 2 2}{8} : \frac{6 + 2}{4}

\frac{2 6 + 2 2}{8}

Factor out common term

2

= \frac{2 ( 6 + 2 )}{8}

Cancel the common factor:

2

= \frac{6 + 2}{4}

= - \frac{6 + 2}{4}

u = \frac{- ( - 2 6 ) - 2 2}{2 ( - 4 )} : - \frac{6 - 2}{4}

\frac{- ( - 2 6 ) - 2 2}{2 ( - 4 )}

Remove parentheses:

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

= \frac{2 6 - 2 2}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 6 - 2 2}{- 8}

Apply the fraction rule:

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

= - \frac{2 6 - 2 2}{8}

Cancel

\frac{2 6 - 2 2}{8} : \frac{6 - 2}{4}

\frac{2 6 - 2 2}{8}

Factor out common term

2

= \frac{2 ( 6 - 2 )}{8}

Cancel the common factor:

2

= \frac{6 - 2}{4}

= - \frac{6 - 2}{4}

The solutions to the quadratic equation are:

u = - \frac{6 + 2}{4}, u = - \frac{6 - 2}{4}

Substitute back

u = cos (x)

cos (x) = - \frac{6 + 2}{4}, cos (x) = - \frac{6 - 2}{4}

cos (x) = - \frac{6 + 2}{4}, cos (x) = - \frac{6 - 2}{4}

cos (x) = - \frac{6 + 2}{4} : x = arccos (- \frac{6 + 2}{4}) + 2 πn, x = - arccos (- \frac{6 + 2}{4}) + 2 πn

cos (x) = - \frac{6 + 2}{4}

Apply trig inverse properties

cos (x) = - \frac{6 + 2}{4}

General solutions for

cos (x) = - \frac{6 + 2}{4}

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

x = arccos (- \frac{6 + 2}{4}) + 2 πn, x = - arccos (- \frac{6 + 2}{4}) + 2 πn

x = arccos (- \frac{6 + 2}{4}) + 2 πn, x = - arccos (- \frac{6 + 2}{4}) + 2 πn

cos (x) = - \frac{6 - 2}{4} : x = arccos (- \frac{6 - 2}{4}) + 2 πn, x = - arccos (- \frac{6 - 2}{4}) + 2 πn

cos (x) = - \frac{6 - 2}{4}

Apply trig inverse properties

cos (x) = - \frac{6 - 2}{4}

General solutions for

cos (x) = - \frac{6 - 2}{4}

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

x = arccos (- \frac{6 - 2}{4}) + 2 πn, x = - arccos (- \frac{6 - 2}{4}) + 2 πn

x = arccos (- \frac{6 - 2}{4}) + 2 πn, x = - arccos (- \frac{6 - 2}{4}) + 2 πn

Combine all the solutions

x = arccos (- \frac{6 + 2}{4}) + 2 πn, x = - arccos (- \frac{6 + 2}{4}) + 2 πn, x = arccos (- \frac{6 - 2}{4}) + 2 πn, x = - arccos (- \frac{6 - 2}{4}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

arccos (- \frac{6 + 2}{4}) + 2 πn :

True

arccos (- \frac{6 + 2}{4}) + 2 πn

Plug in

n = 1

arccos (- \frac{6 + 2}{4}) + 2 π 1

For

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

plug in

x = arccos (- \frac{6 + 2}{4}) + 2 π 1

sin (arccos (- \frac{6 + 2}{4}) + 2 π 1) + 3 cos (arccos (- \frac{6 + 2}{4}) + 2 π 1) = - 2

Refine

- 1.41421 \dots = - 1.41421 \dots

\Rightarrow True

Check the solution

- arccos (- \frac{6 + 2}{4}) + 2 πn :

False

- arccos (- \frac{6 + 2}{4}) + 2 πn

Plug in

n = 1

- arccos (- \frac{6 + 2}{4}) + 2 π 1

For

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

plug in

x = - arccos (- \frac{6 + 2}{4}) + 2 π 1

sin (- arccos (- \frac{6 + 2}{4}) + 2 π 1) + 3 cos (- arccos (- \frac{6 + 2}{4}) + 2 π 1) = - 2

Refine

- 1.93185 \dots = - 1.41421 \dots

\Rightarrow False

Check the solution

arccos (- \frac{6 - 2}{4}) + 2 πn :

False

arccos (- \frac{6 - 2}{4}) + 2 πn

Plug in

n = 1

arccos (- \frac{6 - 2}{4}) + 2 π 1

For

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

plug in

x = arccos (- \frac{6 - 2}{4}) + 2 π 1

sin (arccos (- \frac{6 - 2}{4}) + 2 π 1) + 3 cos (arccos (- \frac{6 - 2}{4}) + 2 π 1) = - 2

Refine

0.51763 \dots = - 1.41421 \dots

\Rightarrow False

Check the solution

- arccos (- \frac{6 - 2}{4}) + 2 πn :

True

- arccos (- \frac{6 - 2}{4}) + 2 πn

Plug in

n = 1

- arccos (- \frac{6 - 2}{4}) + 2 π 1

For

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

plug in

x = - arccos (- \frac{6 - 2}{4}) + 2 π 1

sin (- arccos (- \frac{6 - 2}{4}) + 2 π 1) + 3 cos (- arccos (- \frac{6 - 2}{4}) + 2 π 1) = - 2

Refine

- 1.41421 \dots = - 1.41421 \dots

\Rightarrow True

x = arccos (- \frac{6 + 2}{4}) + 2 πn, x = - arccos (- \frac{6 - 2}{4}) + 2 πn

Show solutions in decimal form

x = 2.87979 \dots + 2 πn, x = - 1.83259 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)+sqrt(3)cos(x)=-sqrt(2) ?
The general solution for sin(x)+sqrt(3)cos(x)=-sqrt(2) is x=2.87979…+2pin,x=-1.83259…+2pin