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

The general solution for (sqrt(3))/2 cos(x)+1/2 sin(x)= 1/2 is x=(11pi)/6+2pin,x= pi/2+2pin

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(sqrt(3))/2 cos(x)+1/2 sin(x)= 1/2

Solution

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

Solution

x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

Degrees

x = 33 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

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

Square both sides

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

Subtract

(\frac{1}{2} - \frac{1}{2} sin (x))^{2}

from both sides

\frac{3}{4} cos^{2} (x) - \frac{1}{4} + \frac{1}{2} sin (x) - \frac{1}{4} sin^{2} (x) = 0

Simplify

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

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

\frac{3}{4} cos^{2} (x) = \frac{3 cos ^{2} ( x )}{4}

\frac{3}{4} cos^{2} (x)

Multiply fractions:

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

= \frac{3 cos ^{2} ( x )}{4}

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

\frac{1}{4} sin^{2} (x) = \frac{sin ^{2} ( x )}{4}

\frac{1}{4} sin^{2} (x)

Multiply fractions:

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

= \frac{1 \cdot sin ^{2} ( x )}{4}

Multiply:

1 \cdot sin^{2} (x) = sin^{2} (x)

= \frac{sin ^{2} ( x )}{4}

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

Combine the fractions

\frac{3 cos ^{2} ( x )}{4} - \frac{1}{4} - \frac{sin ^{2} ( x )}{4} : \frac{3 cos ^{2} ( x ) - 1 - sin ^{2} ( x )}{4}

Apply rule

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

= \frac{3 cos ^{2} ( x ) - 1 - sin ^{2} ( x )}{4}

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

Least Common Multiplier of

4, 2 : 4

4, 2

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

4

2

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{sin ( x )}{2} :

multiply the denominator and numerator by

2

\frac{sin ( x )}{2} = \frac{sin ( x ) \cdot 2}{2 \cdot 2} = \frac{sin ( x ) \cdot 2}{4}

= \frac{3 cos ^{2} ( x ) - 1 - sin ^{2} ( x )}{4} + \frac{sin ( x ) \cdot 2}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{3 cos ^{2} ( x ) - 1 - sin ^{2} ( x ) + sin ( x ) \cdot 2}{4}

\frac{3 cos ^{2} ( x ) - 1 - sin ^{2} ( x ) + 2 sin ( x )}{4} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

3 (1 - sin^{2} (x)) : 3 - 3 sin^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (x)

= 3 \cdot 1 - 3 sin^{2} (x)

Multiply the numbers:

3 \cdot 1 = 3

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add/Subtract the numbers:

- 1 + 3 = 2

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

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 )} : - \frac{1}{2}

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

Remove parentheses:

(- a) = - a

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

Add/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}

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

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

Remove parentheses:

(- a) = - a

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

Subtract 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

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

General solutions for

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

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

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

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

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

plug in

x = \frac{7 π}{6} + 2 π 1

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

Refine

- 1 = 0.5

\Rightarrow False

Check the solution

\frac{11 π}{6} + 2 πn :

True

\frac{11 π}{6} + 2 πn

Plug in

n = 1

\frac{11 π}{6} + 2 π 1

For

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

plug in

x = \frac{11 π}{6} + 2 π 1

\frac{3}{2} cos (\frac{11 π}{6} + 2 π 1) + \frac{1}{2} sin (\frac{11 π}{6} + 2 π 1) = \frac{1}{2}

Refine

0.5 = 0.5

\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

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

plug in

x = \frac{π}{2} + 2 π 1

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

Refine

0.5 = 0.5

\Rightarrow True

x = \frac{11 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))/2 cos(x)+1/2 sin(x)= 1/2 ?
The general solution for (sqrt(3))/2 cos(x)+1/2 sin(x)= 1/2 is x=(11pi)/6+2pin,x= pi/2+2pin