What is the general solution for sin(x+10)=-tan(15) ?

The general solution for sin(x+10)=-tan(15) is No Solution for x\in\mathbb{R}

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Popular >

sin(x+10)=-tan(15)

Solution

sin (x + 1 0^{\circ}) = - tan (1 5^{\circ})

Solution

No Solution for x \in R

Solution steps

sin (x + 1 0^{\circ}) = - tan (1 5^{\circ})

tan (1 5^{\circ}) = 2 - 3

tan (1 5^{\circ})

Rewrite using trig identities:

\frac{tan ( 4 5 ^{\circ} ) - tan ( 3 0 ^{\circ} )}{1 + tan ( 4 5 ^{\circ} ) tan ( 3 0 ^{\circ} )}

tan (1 5^{\circ})

Write

tan (1 5^{\circ})

tan (4 5^{\circ} - 3 0^{\circ})

= tan (4 5^{\circ} - 3 0^{\circ})

Use the Angle Difference identity:

tan (s - t) = \frac{tan ( s ) - tan ( t )}{1 + tan ( s ) tan ( t )}

= \frac{tan ( 4 5 ^{\circ} ) - tan ( 3 0 ^{\circ} )}{1 + tan ( 4 5 ^{\circ} ) tan ( 3 0 ^{\circ} )}

= \frac{tan ( 4 5 ^{\circ} ) - tan ( 3 0 ^{\circ} )}{1 + tan ( 4 5 ^{\circ} ) tan ( 3 0 ^{\circ} )}

Use the following trivial identity:

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

tan (x)

periodicity table with

18 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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

Use the following trivial identity:

tan (3 0^{\circ}) = \frac{3}{3}

tan (3 0^{\circ})

tan (x)

periodicity table with

18 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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= \frac{3}{3}

= \frac{1 - \frac{3}{3}}{1 + 1 \cdot \frac{3}{3}}

Simplify

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

\frac{1 - \frac{3}{3}}{1 + 1 \cdot \frac{3}{3}}

Multiply:

1 \cdot \frac{3}{3} = \frac{3}{3}

= \frac{1 - \frac{3}{3}}{1 + \frac{3}{3}}

Join

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

1 + \frac{3}{3}

Convert element to fraction:

1 = \frac{1 \cdot 3}{3}

= \frac{1 \cdot 3}{3} + \frac{3}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 3 + 3}{3}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3 + 3}{3}

Factor

3 + 3 : 3 (3 + 1)

3 + 3

3 = 33

= 33 + 3

Factor out common term

3

= 3 (3 + 1)

= \frac{3 ( 3 + 1 )}{3}

Cancel

\frac{3 ( 3 + 1 )}{3} : \frac{3 + 1}{3}

\frac{3 ( 3 + 1 )}{3}

Apply radical rule:

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}} ( 1 + 3 )}{3}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{3 + 1}{3 ^{1 - \frac{1}{2}}}

Subtract the numbers:

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

= \frac{3 + 1}{3 ^{\frac{1}{2}}}

Apply radical rule:

3^{\frac{1}{2}} = 3

= \frac{3 + 1}{3}

= \frac{3 + 1}{3}

= \frac{1 - \frac{3}{3}}{\frac{3 + 1}{3}}

Join

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

1 - \frac{3}{3}

Convert element to fraction:

1 = \frac{1 \cdot 3}{3}

= \frac{1 \cdot 3}{3} - \frac{3}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 3 - 3}{3}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3 - 3}{3}

Factor

3 - 3 : 3 (3 - 1)

3 - 3

3 = 33

= 33 - 3

Factor out common term

3

= 3 (3 - 1)

= \frac{3 ( 3 - 1 )}{3}

Cancel

\frac{3 ( 3 - 1 )}{3} : \frac{3 - 1}{3}

\frac{3 ( 3 - 1 )}{3}

Apply radical rule:

3 = 3^{\frac{1}{2}}

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{3 - 1}{3 ^{1 - \frac{1}{2}}}

Subtract the numbers:

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

= \frac{3 - 1}{3 ^{\frac{1}{2}}}

Apply radical rule:

3^{\frac{1}{2}} = 3

= \frac{3 - 1}{3}

= \frac{3 - 1}{3}

= \frac{\frac{3 - 1}{3}}{\frac{3 + 1}{3}}

Divide fractions:

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

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

Cancel the common factor:

3

= \frac{3 - 1}{3 + 1}

Rationalize

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

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

Multiply by the conjugate

\frac{3 - 1}{3 - 1}

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

(3 - 1) (3 - 1) = 4 - 23

(3 - 1) (3 - 1)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

(3 - 1) (3 - 1) = (3 - 1)^{1 + 1}

= (3 - 1)^{1 + 1}

Add the numbers:

1 + 1 = 2

= (3 - 1)^{2}

Apply Perfect Square Formula:

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

a = 3, b = 1

= (3)^{2} - 23 \cdot 1 + 1^{2}

Simplify

(3)^{2} - 23 \cdot 1 + 1^{2} : 4 - 23

(3)^{2} - 23 \cdot 1 + 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

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

(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

23 \cdot 1 = 23

23 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 23

= 3 - 23 + 1

Add the numbers:

3 + 1 = 4

= 4 - 23

= 4 - 23

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

(3 + 1) (3 - 1)

Apply Difference of Two Squares Formula:

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

a = 3, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= (3)^{2} - 1

(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

= 3 - 1

Subtract the numbers:

3 - 1 = 2

= 2

= 2

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

Factor

4 - 23 : 2 (2 - 3)

4 - 23

Rewrite as

= 2 \cdot 2 - 23

Factor out common term

2

= 2 (2 - 3)

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

Divide the numbers:

\frac{2}{2} = 1

= 2 - 3

= 2 - 3

= 2 - 3

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

- 1 \leq sin (x) \leq 1

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x+10)=-tan(15) ?
The general solution for sin(x+10)=-tan(15) is No Solution for x\in\mathbb{R}