What is the general solution for sqrt(3)sin(x)=1+cos(x) ?

The general solution for sqrt(3)sin(x)=1+cos(x) is x=pi+2pin,x= pi/3+2pin

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sqrt(3)sin(x)=1+cos(x)

Solution

3 sin (x) = 1 + cos (x)

Solution

x = π + 2 πn, x = \frac{π}{3} + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (x) = 1 + cos (x)

Square both sides

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

Subtract

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

from both sides

3 sin^{2} (x) - 1 - 2 cos (x) - cos^{2} (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add/Subtract the numbers:

- 1 + 3 = 2

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

Add the numbers:

4 + 32 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

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

Separate the solutions

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

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

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

Remove parentheses:

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

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

Add the numbers:

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

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

Remove parentheses:

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

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

Subtract the numbers:

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

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

3 sin (x) = 1 + cos (x)

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

3 sin (x) = 1 + cos (x)

plug in

x = π + 2 π 1

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

Refine

0 = 0

\Rightarrow True

Check the solution

\frac{π}{3} + 2 πn :

True

\frac{π}{3} + 2 πn

Plug in

n = 1

\frac{π}{3} + 2 π 1

For

3 sin (x) = 1 + cos (x)

plug in

x = \frac{π}{3} + 2 π 1

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

Refine

1.5 = 1.5

\Rightarrow True

Check the solution

\frac{5 π}{3} + 2 πn :

False

\frac{5 π}{3} + 2 πn

Plug in

n = 1

\frac{5 π}{3} + 2 π 1

For

3 sin (x) = 1 + cos (x)

plug in

x = \frac{5 π}{3} + 2 π 1

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

Refine

- 1.5 = 1.5

\Rightarrow False

x = π + 2 πn, x = \frac{π}{3} + 2 πn

Popular Examples

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sin (\frac{1}{2} x) = 0

cos^2(x)-2cos(x)-1=0

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

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\frac{sin ( x )}{cos ( x )} = 1

9(1-cos(θ))=sin^2(θ)

9 (1 - cos (θ)) = sin^{2} (θ)

solvefor x,1=sin(x)solve for

x, 1 = sin (x)

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)sin(x)=1+cos(x) ?
The general solution for sqrt(3)sin(x)=1+cos(x) is x=pi+2pin,x= pi/3+2pin