What is the general solution for sin(x-30)cos(x-30)=(sqrt(3))/4 ?

The general solution for sin(x-30)cos(x-30)=(sqrt(3))/4 is x=60+180n,x=180n+90

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sin(x-30)cos(x-30)=(sqrt(3))/4

Solution

sin (x - 3 0^{\circ}) cos (x - 3 0^{\circ}) = \frac{3}{4}

Solution

x = 6 0^{\circ} + 18 0^{\circ} n, x = 18 0^{\circ} n + 9 0^{\circ}

Radians

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

Solution steps

sin (x - 3 0^{\circ}) cos (x - 3 0^{\circ}) = \frac{3}{4}

Rewrite using trig identities

sin (x - 3 0^{\circ}) cos (x - 3 0^{\circ})

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

= \frac{sin ( 2 ( - 3 0 ^{\circ} + x ) )}{2}

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

Multiply both sides by

2

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

Multiply both sides by

2

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

Simplify

sin (2 (- 3 0^{\circ} + x)) = \frac{3}{2}

sin (2 (- 3 0^{\circ} + x)) = \frac{3}{2}

General solutions for

sin (2 (- 3 0^{\circ} + x)) = \frac{3}{2}

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 (- 3 0^{\circ} + x) = 6 0^{\circ} + 36 0^{\circ} n, 2 (- 3 0^{\circ} + x) = 12 0^{\circ} + 36 0^{\circ} n

2 (- 3 0^{\circ} + x) = 6 0^{\circ} + 36 0^{\circ} n, 2 (- 3 0^{\circ} + x) = 12 0^{\circ} + 36 0^{\circ} n

Solve

2 (- 3 0^{\circ} + x) = 6 0^{\circ} + 36 0^{\circ} n : x = 6 0^{\circ} + 18 0^{\circ} n

2 (- 3 0^{\circ} + x) = 6 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 (- 3 0^{\circ} + x) = 6 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 ( - 3 0 ^{\circ} + x )}{2} = \frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 ( - 3 0 ^{\circ} + x )}{2} = \frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 ( - 3 0 ^{\circ} + x )}{2} : - 3 0^{\circ} + x

\frac{2 ( - 3 0 ^{\circ} + x )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 3 0^{\circ} + x

Simplify

\frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 3 0^{\circ} + 18 0^{\circ} n

\frac{6 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{6 0 ^{\circ}}{2} = 3 0^{\circ}

\frac{6 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{18 0 ^{\circ}}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= 3 0^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 3 0^{\circ} + 18 0^{\circ} n

- 3 0^{\circ} + x = 3 0^{\circ} + 18 0^{\circ} n

- 3 0^{\circ} + x = 3 0^{\circ} + 18 0^{\circ} n

- 3 0^{\circ} + x = 3 0^{\circ} + 18 0^{\circ} n

Move

3 0^{\circ}

to the right side

- 3 0^{\circ} + x = 3 0^{\circ} + 18 0^{\circ} n

Add

3 0^{\circ}

to both sides

- 3 0^{\circ} + x + 3 0^{\circ} = 3 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ}

Simplify

- 3 0^{\circ} + x + 3 0^{\circ} = 3 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ}

Simplify

- 3 0^{\circ} + x + 3 0^{\circ} : x

- 3 0^{\circ} + x + 3 0^{\circ}

Add similar elements:

- 3 0^{\circ} + 3 0^{\circ} = 0

= x

Simplify

3 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ} : 6 0^{\circ} + 18 0^{\circ} n

3 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ}

Group like terms

= 3 0^{\circ} + 3 0^{\circ} + 18 0^{\circ} n

Combine the fractions

3 0^{\circ} + 3 0^{\circ} : 6 0^{\circ}

Apply rule

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

= \frac{18 0 ^{\circ} + 18 0 ^{\circ}}{6}

Add similar elements:

18 0^{\circ} + 18 0^{\circ} = 36 0^{\circ}

= 6 0^{\circ}

Cancel the common factor:

2

= 6 0^{\circ}

= 6 0^{\circ} + 18 0^{\circ} n

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

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

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

Solve

2 (- 3 0^{\circ} + x) = 12 0^{\circ} + 36 0^{\circ} n : x = 18 0^{\circ} n + 9 0^{\circ}

2 (- 3 0^{\circ} + x) = 12 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 (- 3 0^{\circ} + x) = 12 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 ( - 3 0 ^{\circ} + x )}{2} = \frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 ( - 3 0 ^{\circ} + x )}{2} = \frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 ( - 3 0 ^{\circ} + x )}{2} : - 3 0^{\circ} + x

\frac{2 ( - 3 0 ^{\circ} + x )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 3 0^{\circ} + x

Simplify

\frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 6 0^{\circ} + 18 0^{\circ} n

\frac{12 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{12 0 ^{\circ}}{2} = 6 0^{\circ}

\frac{12 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{36 0 ^{\circ}}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= 6 0^{\circ}

Cancel the common factor:

2

= 6 0^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 6 0^{\circ} + 18 0^{\circ} n

- 3 0^{\circ} + x = 6 0^{\circ} + 18 0^{\circ} n

- 3 0^{\circ} + x = 6 0^{\circ} + 18 0^{\circ} n

- 3 0^{\circ} + x = 6 0^{\circ} + 18 0^{\circ} n

Move

3 0^{\circ}

to the right side

- 3 0^{\circ} + x = 6 0^{\circ} + 18 0^{\circ} n

Add

3 0^{\circ}

to both sides

- 3 0^{\circ} + x + 3 0^{\circ} = 6 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ}

Simplify

- 3 0^{\circ} + x + 3 0^{\circ} = 6 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ}

Simplify

- 3 0^{\circ} + x + 3 0^{\circ} : x

- 3 0^{\circ} + x + 3 0^{\circ}

Add similar elements:

- 3 0^{\circ} + 3 0^{\circ} = 0

= x

Simplify

6 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ} : 18 0^{\circ} n + 9 0^{\circ}

6 0^{\circ} + 18 0^{\circ} n + 3 0^{\circ}

Group like terms

= 18 0^{\circ} n + 6 0^{\circ} + 3 0^{\circ}

Least Common Multiplier of

3, 6 : 6

3, 6

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

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

3

6

= 3 \cdot 2

Multiply the numbers:

3 \cdot 2 = 6

= 6

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

6

For

6 0^{\circ} :

multiply the denominator and numerator by

2

6 0^{\circ} = \frac{18 0 ^{\circ} 2}{3 \cdot 2} = 6 0^{\circ}

= 6 0^{\circ} + 3 0^{\circ}

Since the denominators are equal, combine the fractions:

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

= \frac{18 0 ^{\circ} 2 + 18 0 ^{\circ}}{6}

Add similar elements:

36 0^{\circ} + 18 0^{\circ} = 54 0^{\circ}

= 9 0^{\circ}

Cancel the common factor:

3

= 18 0^{\circ} n + 9 0^{\circ}

x = 18 0^{\circ} n + 9 0^{\circ}

x = 18 0^{\circ} n + 9 0^{\circ}

x = 18 0^{\circ} n + 9 0^{\circ}

x = 6 0^{\circ} + 18 0^{\circ} n, x = 18 0^{\circ} n + 9 0^{\circ}

Popular Examples

cos(θ)=-7/15 ,cos(θ/2),180<θ<270

cos (θ) = - \frac{7}{15}, cos (\frac{θ}{2}), 18 0^{\circ} < θ < 27 0^{\circ}

tan(x)+2=0

tan (x) + 2 = 0

26^2=13^2+19^2-(13)(19)cos(J)

2 6^{2} = 1 3^{2} + 1 9^{2} - (13) (19) cos (J)

4=sec^2(x)

4 = sec^{2} (x)

tan(x/2)=4

tan (\frac{x}{2}) = 4

Frequently Asked Questions (FAQ)

What is the general solution for sin(x-30)cos(x-30)=(sqrt(3))/4 ?
The general solution for sin(x-30)cos(x-30)=(sqrt(3))/4 is x=60+180n,x=180n+90