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

The general solution for sin(x)-sqrt(3)cos(x)=sqrt(3) is x=pi+2pin,x=(2pi)/3+2pin

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

Solution

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

Solution

x = π + 2 πn, x = \frac{2 π}{3} + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Add

3 cos (x)

to both sides

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

Square both sides

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

Subtract

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

from both sides

sin^{2} (x) - 3 - 6 cos (x) - 3 cos^{2} (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Add similar elements:

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

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

Add/Subtract the numbers:

- 3 + 1 = - 2

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 4, b = - 6, c = - 2

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

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

(- 6)^{2} - 4 (- 4) (- 2) = 2

(- 6)^{2} - 4 (- 4) (- 2)

Apply rule

- (- a) = a

= (- 6)^{2} - 4 \cdot 4 \cdot 2

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 6^{2} - 32

6^{2} = 36

= 36 - 32

Subtract the numbers:

36 - 32 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

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

Separate the solutions

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

u = \frac{- ( - 6 ) + 2}{2 ( - 4 )} : - 1

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

Remove parentheses:

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

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

Add the numbers:

6 + 2 = 8

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{- 8}

Apply the fraction rule:

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

= - \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 6 ) - 2}{2 ( - 4 )} : - \frac{1}{2}

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

Remove parentheses:

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

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

Subtract the numbers:

6 - 2 = 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{- 8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = - 1, u = - \frac{1}{2}

Substitute back

u = cos (x)

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

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

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

General solutions for

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

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

x = π + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

π + 2 πn :

True

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

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

plug in

x = π + 2 π 1

sin (π + 2 π 1) - 3 cos (π + 2 π 1) = 3

Refine

1.73205 \dots = 1.73205 \dots

\Rightarrow True

Check the solution

\frac{2 π}{3} + 2 πn :

True

\frac{2 π}{3} + 2 πn

Plug in

n = 1

\frac{2 π}{3} + 2 π 1

For

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

plug in

x = \frac{2 π}{3} + 2 π 1

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

Refine

1.73205 \dots = 1.73205 \dots

\Rightarrow True

Check the solution

\frac{4 π}{3} + 2 πn :

False

\frac{4 π}{3} + 2 πn

Plug in

n = 1

\frac{4 π}{3} + 2 π 1

For

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

plug in

x = \frac{4 π}{3} + 2 π 1

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

Refine

0 = 1.73205 \dots

\Rightarrow False

x = π + 2 πn, x = \frac{2 π}{3} + 2 πn

Popular Examples

cot^2(x)-2csc(x)=-1

cot^{2} (x) - 2 csc (x) = - 1

sqrt(3)csc(x)+csc(x)cot(x)=0

3 csc (x) + csc (x) cot (x) = 0

-3cos^2(θ)-10cos(θ)=-3cos(θ)-6

- 3 cos^{2} (θ) - 10 cos (θ) = - 3 cos (θ) - 6

-4sin(x)=-cos^2(x)+4

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

sin^2(x)-4sin(x)+3=0

sin^{2} (x) - 4 sin (x) + 3 = 0

Frequently Asked Questions (FAQ)

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