What is the general solution for sin(x)=sin(x+30) ?

The general solution for sin(x)=sin(x+30) is x=1.30899…+180n

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Popular >

sin(x)=sin(x+30)

Solution

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

Solution

x = 1.30899 \dots + 18 0^{\circ} n

Radians

x = 1.30899 \dots + πn

Solution steps

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

Rewrite using trig identities

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

Rewrite using trig identities

sin (x + 3 0^{\circ})

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

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

Simplify

sin (x) cos (3 0^{\circ}) + cos (x) sin (3 0^{\circ}) : \frac{3}{2} sin (x) + \frac{1}{2} cos (x)

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

Simplify

cos (3 0^{\circ}) : \frac{3}{2}

cos (3 0^{\circ})

Use the following trivial identity:

cos (3 0^{\circ}) = \frac{3}{2}

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

= \frac{3}{2}

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

Simplify

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

Use the following trivial identity:

sin (3 0^{\circ}) = \frac{1}{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}

= \frac{1}{2}

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

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

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

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

Subtract

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

from both sides

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

Simplify

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

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

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

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

Multiply fractions:

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

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

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

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

Apply rule

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

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

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

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

(2 - 3) sin (x) - cos (x) = 0

Rewrite using trig identities

(2 - 3) sin (x) - cos (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

(2 - 3) tan (x) - 1 = 0

(2 - 3) tan (x) - 1 = 0

Move

1

to the right side

(2 - 3) tan (x) - 1 = 0

Add

1

to both sides

(2 - 3) tan (x) - 1 + 1 = 0 + 1

Simplify

(2 - 3) tan (x) = 1

(2 - 3) tan (x) = 1

Divide both sides by

2 - 3

(2 - 3) tan (x) = 1

Divide both sides by

2 - 3

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

Simplify

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

Simplify

\frac{( 2 - 3 ) tan ( x )}{2 - 3} : tan (x)

\frac{( 2 - 3 ) tan ( x )}{2 - 3}

Cancel the common factor:

2 - 3

= tan (x)

Simplify

\frac{1}{2 - 3} : 2 + 3

\frac{1}{2 - 3}

Multiply by the conjugate

\frac{2 + 3}{2 + 3}

= \frac{1 \cdot ( 2 + 3 )}{( 2 - 3 ) ( 2 + 3 )}

1 \cdot (2 + 3) = 2 + 3

(2 - 3) (2 + 3) = 1

(2 - 3) (2 + 3)

Apply Difference of Two Squares Formula:

(a - b) (a + b) = a^{2} - b^{2}

a = 2, b = 3

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

Simplify

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

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

2^{2} = 4

2^{2}

2^{2} = 4

= 4

(3)^{2} = 3

(3)^{2}

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 - 3

Subtract the numbers:

4 - 3 = 1

= 1

= 1

= \frac{2 + 3}{1}

Apply rule

\frac{a}{1} = a

= 2 + 3

tan (x) = 2 + 3

tan (x) = 2 + 3

tan (x) = 2 + 3

Apply trig inverse properties

tan (x) = 2 + 3

General solutions for

tan (x) = 2 + 3

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

x = arctan (2 + 3) + 18 0^{\circ} n

x = arctan (2 + 3) + 18 0^{\circ} n

Show solutions in decimal form

x = 1.30899 \dots + 18 0^{\circ} n

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)=sin(x+30) ?
The general solution for sin(x)=sin(x+30) is x=1.30899…+180n