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

The general solution for sin(x)-sqrt(3)cos(x)=2 is x=(5pi)/6+2pin

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

Solution

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

Solution

x = \frac{5 π}{6} + 2 πn

Degrees

x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Add

3 cos (x)

to 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) - 4 - 43 cos (x) - 3 cos^{2} (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- 4 + 1 - cos^{2} (x) - 3 cos^{2} (x) - 4 cos (x) 3 : - 4 cos^{2} (x) - 43 cos (x) - 3

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

Add similar elements:

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

= - 4 + 1 - 4 cos^{2} (x) - 43 cos (x)

Add/Subtract the numbers:

- 4 + 1 = - 3

= - 4 cos^{2} (x) - 43 cos (x) - 3

= - 4 cos^{2} (x) - 43 cos (x) - 3

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

Solve by substitution

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

Let:

cos (x) = u

- 3 - 4 u^{2} - 4 u 3 = 0

- 3 - 4 u^{2} - 4 u 3 = 0 : u = - \frac{3}{2}

- 3 - 4 u^{2} - 4 u 3 = 0

Write in the standard form

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

- 4 u^{2} - 43 u - 3 = 0

Solve with the quadratic formula

- 4 u^{2} - 43 u - 3 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 43, c = - 3

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

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

(- 43)^{2} - 4 (- 4) (- 3) = 0

(- 43)^{2} - 4 (- 4) (- 3)

Apply rule

- (- a) = a

= (- 43)^{2} - 4 \cdot 4 \cdot 3

(- 43)^{2} = 4^{2} \cdot 3

(- 43)^{2}

Apply exponent rule:

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

n

is even

(- 43)^{2} = (43)^{2}

= (43)^{2}

Apply exponent rule:

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

= 4^{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

= 4^{2} \cdot 3

4 \cdot 4 \cdot 3 = 48

4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 48

= 4^{2} \cdot 3 - 48

4^{2} \cdot 3 = 48

4^{2} \cdot 3

4^{2} = 16

= 16 \cdot 3

Multiply the numbers:

16 \cdot 3 = 48

= 48

= 48 - 48

Subtract the numbers:

48 - 48 = 0

= 0

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

u = \frac{- ( - 4 3 )}{2 ( - 4 )}

\frac{- ( - 4 3 )}{2 ( - 4 )} = - \frac{3}{2}

\frac{- ( - 4 3 )}{2 ( - 4 )}

Remove parentheses:

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

= \frac{4 3}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4 3}{- 8}

Apply the fraction rule:

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

= - \frac{4 3}{8}

Cancel the common factor:

4

= - \frac{3}{2}

u = - \frac{3}{2}

The solution to the quadratic equation is:

u = - \frac{3}{2}

Substitute back

u = cos (x)

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

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

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

Combine all the solutions

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

\frac{5 π}{6} + 2 πn :

True

\frac{5 π}{6} + 2 πn

Plug in

n = 1

\frac{5 π}{6} + 2 π 1

For

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

plug in

x = \frac{5 π}{6} + 2 π 1

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

Refine

2 = 2

\Rightarrow True

Check the solution

\frac{7 π}{6} + 2 πn :

False

\frac{7 π}{6} + 2 πn

Plug in

n = 1

\frac{7 π}{6} + 2 π 1

For

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

plug in

x = \frac{7 π}{6} + 2 π 1

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

Refine

1 = 2

\Rightarrow False

x = \frac{5 π}{6} + 2 πn

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3 tan (15 x^{5}) + 1 = 0

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arctan(θ)=sqrt(3)

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)-sqrt(3)cos(x)=2 ?
The general solution for sin(x)-sqrt(3)cos(x)=2 is x=(5pi)/6+2pin