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

The general solution for sqrt(3)sin(x)-3cos(x)=sqrt(3) is x=(7pi)/6+2pin,x= pi/2+2pin

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

Solution

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

Solution

x = \frac{7 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

Degrees

x = 21 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Add

3 cos (x)

to both sides

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

Square both sides

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

Subtract

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

from both sides

3 sin^{2} (x) - 3 - 63 cos (x) - 9 cos^{2} (x) = 0

Rewrite using trig identities

- 3 + 3 sin^{2} (x) - 9 cos^{2} (x) - 6 cos (x) 3

Use the Pythagorean identity:

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

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

= - 3 + 3 (1 - cos^{2} (x)) - 9 cos^{2} (x) - 6 cos (x) 3

Simplify

- 3 + 3 (1 - cos^{2} (x)) - 9 cos^{2} (x) - 6 cos (x) 3 : - 12 cos^{2} (x) - 63 cos (x)

- 3 + 3 (1 - cos^{2} (x)) - 9 cos^{2} (x) - 6 cos (x) 3

= - 3 + 3 (1 - cos^{2} (x)) - 9 cos^{2} (x) - 63 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)

= - 3 + 3 - 3 cos^{2} (x) - 9 cos^{2} (x) - 6 cos (x) 3

Simplify

- 3 + 3 - 3 cos^{2} (x) - 9 cos^{2} (x) - 6 cos (x) 3 : - 12 cos^{2} (x) - 63 cos (x)

- 3 + 3 - 3 cos^{2} (x) - 9 cos^{2} (x) - 6 cos (x) 3

Add similar elements:

- 3 cos^{2} (x) - 9 cos^{2} (x) = - 12 cos^{2} (x)

= - 3 + 3 - 12 cos^{2} (x) - 63 cos (x)

- 3 + 3 = 0

= - 12 cos^{2} (x) - 63 cos (x)

= - 12 cos^{2} (x) - 63 cos (x)

= - 12 cos^{2} (x) - 63 cos (x)

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

Solve by substitution

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

Let:

cos (x) = u

- 12 u^{2} - 6 u 3 = 0

- 12 u^{2} - 6 u 3 = 0 : u = - \frac{3}{2}, u = 0

- 12 u^{2} - 6 u 3 = 0

Write in the standard form

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

- 12 u^{2} - 63 u = 0

Solve with the quadratic formula

- 12 u^{2} - 63 u = 0

Quadratic Equation Formula:

For

a = - 12, b = - 63, c = 0

u_{1, 2} = \frac{- ( - 6 3 ) \pm ( - 6 3 ) ^{2} - 4 ( - 12 ) \cdot 0}{2 ( - 12 )}

u_{1, 2} = \frac{- ( - 6 3 ) \pm ( - 6 3 ) ^{2} - 4 ( - 12 ) \cdot 0}{2 ( - 12 )}

(- 63)^{2} - 4 (- 12) \cdot 0 = 63

(- 63)^{2} - 4 (- 12) \cdot 0

Apply rule

- (- a) = a

= (- 63)^{2} + 4 \cdot 12 \cdot 0

(- 63)^{2} = 6^{2} \cdot 3

(- 63)^{2}

Apply exponent rule:

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

n

is even

(- 63)^{2} = (63)^{2}

= (63)^{2}

Apply exponent rule:

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

= 6^{2} (3)^{2}

(3)^{2} : 3

Apply radical rule:

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

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 3^{\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

= 3

= 6^{2} \cdot 3

4 \cdot 12 \cdot 0 = 0

4 \cdot 12 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 6^{2} \cdot 3 + 0

6^{2} \cdot 3 + 0 = 6^{2} \cdot 3

= 6^{2} \cdot 3

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 3 6^{2}

Apply radical rule:

assuming

a \geq 0

6^{2} = 6

= 63

u_{1, 2} = \frac{- ( - 6 3 ) \pm 6 3}{2 ( - 12 )}

Separate the solutions

u_{1} = \frac{- ( - 6 3 ) + 6 3}{2 ( - 12 )}, u_{2} = \frac{- ( - 6 3 ) - 6 3}{2 ( - 12 )}

u = \frac{- ( - 6 3 ) + 6 3}{2 ( - 12 )} : - \frac{3}{2}

\frac{- ( - 6 3 ) + 6 3}{2 ( - 12 )}

Remove parentheses:

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

= \frac{6 3 + 6 3}{- 2 \cdot 12}

Add similar elements:

63 + 63 = 123

= \frac{12 3}{- 2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{12 3}{- 24}

Apply the fraction rule:

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

= - \frac{12 3}{24}

Cancel the common factor:

12

= - \frac{3}{2}

u = \frac{- ( - 6 3 ) - 6 3}{2 ( - 12 )} : 0

\frac{- ( - 6 3 ) - 6 3}{2 ( - 12 )}

Remove parentheses:

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

= \frac{6 3 - 6 3}{- 2 \cdot 12}

Add similar elements:

63 - 63 = 0

= \frac{0}{- 2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{0}{- 24}

Apply the fraction rule:

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

= - \frac{0}{24}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

The solutions to the quadratic equation are:

u = - \frac{3}{2}, u = 0

Substitute back

u = cos (x)

cos (x) = - \frac{3}{2}, cos (x) = 0

cos (x) = - \frac{3}{2}, cos (x) = 0

cos (x) = - \frac{3}{2} : x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

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

General solutions for

cos (x) = - \frac{3}{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{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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

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

Combine all the solutions

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

Check the solution

\frac{5 π}{6} + 2 πn :

False

\frac{5 π}{6} + 2 πn

Plug in

n = 1

\frac{5 π}{6} + 2 π 1

For

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

plug in

x = \frac{5 π}{6} + 2 π 1

3 sin (\frac{5 π}{6} + 2 π 1) - 3 cos (\frac{5 π}{6} + 2 π 1) = 3

Refine

3.46410 \dots = 1.73205 \dots

\Rightarrow False

Check the solution

\frac{7 π}{6} + 2 πn :

True

\frac{7 π}{6} + 2 πn

Plug in

n = 1

\frac{7 π}{6} + 2 π 1

For

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

plug in

x = \frac{7 π}{6} + 2 π 1

3 sin (\frac{7 π}{6} + 2 π 1) - 3 cos (\frac{7 π}{6} + 2 π 1) = 3

Refine

1.73205 \dots = 1.73205 \dots

\Rightarrow True

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

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

plug in

x = \frac{π}{2} + 2 π 1

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

Refine

1.73205 \dots = 1.73205 \dots

\Rightarrow True

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

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

plug in

x = \frac{3 π}{2} + 2 π 1

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

Refine

- 1.73205 \dots = 1.73205 \dots

\Rightarrow False

x = \frac{7 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)sin(x)-3cos(x)=sqrt(3) ?
The general solution for sqrt(3)sin(x)-3cos(x)=sqrt(3) is x=(7pi)/6+2pin,x= pi/2+2pin