What is the general solution for sec(40+2θ)=csc(15) ?

The general solution for sec(40+2θ)=csc(15) is θ=180n-20+(1.30899…)/2 ,θ=180+180n-20-(1.30899…)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sec(40+2θ)=csc(15)

Solution

sec (4 0^{\circ} + 2 θ) = csc (1 5^{\circ})

Solution

θ = 18 0^{\circ} n - 2 0^{\circ} + \frac{1.30899 \dots}{2}, θ = 18 0^{\circ} + 18 0^{\circ} n - 2 0^{\circ} - \frac{1.30899 \dots}{2}

Radians

θ = - \frac{π}{9} + \frac{1.30899 \dots}{2} + πn, θ = π - \frac{π}{9} - \frac{1.30899 \dots}{2} + πn

Solution steps

sec (4 0^{\circ} + 2 θ) = csc (1 5^{\circ})

csc (1 5^{\circ}) = 6 + 2

csc (1 5^{\circ})

Rewrite using trig identities:

\frac{1}{sin ( 1 5 ^{\circ} )}

csc (1 5^{\circ})

Use the basic trigonometric identity:

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

= \frac{1}{sin ( 1 5 ^{\circ} )}

= \frac{1}{sin ( 1 5 ^{\circ} )}

Rewrite using trig identities:

sin (1 5^{\circ}) = \frac{6 - 2}{4}

sin (1 5^{\circ})

Rewrite using trig identities:

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

sin (1 5^{\circ})

Write

sin (1 5^{\circ})

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

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

Use the Angle Difference identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

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

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

Use the following trivial identity:

sin (4 5^{\circ}) = \frac{2}{2}

sin (4 5^{\circ})

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{2}{2}

Use the following trivial identity:

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

cos (3 0^{\circ})

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}

Use the following trivial identity:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\circ})

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{2}{2}

Use the following trivial identity:

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

sin (3 0^{\circ})

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

Simplify

\frac{2}{2} \cdot \frac{3}{2} - \frac{2}{2} \cdot \frac{1}{2} : \frac{6 - 2}{4}

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

\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}

\frac{2}{2} \cdot \frac{3}{2}

Multiply fractions:

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

= \frac{2 3}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 3}{4}

Simplify

23 : 6

23

Apply radical rule:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

\frac{2}{2} \cdot \frac{1}{2} = \frac{2}{4}

\frac{2}{2} \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{2 \cdot 1}{2 \cdot 2}

Multiply:

2 \cdot 1 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

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

Apply rule

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

= \frac{6 - 2}{4}

= \frac{6 - 2}{4}

= \frac{1}{\frac{6 - 2}{4}}

Simplify

\frac{1}{\frac{6 - 2}{4}} : 6 + 2

\frac{1}{\frac{6 - 2}{4}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{4}{6 - 2}

Rationalize

\frac{4}{6 - 2} : 6 + 2

\frac{4}{6 - 2}

Multiply by the conjugate

\frac{6 + 2}{6 + 2}

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

(6 - 2) (6 + 2) = 4

(6 - 2) (6 + 2)

Apply Difference of Two Squares Formula:

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

a = 6, b = 2

= (6)^{2} - (2)^{2}

Simplify

(6)^{2} - (2)^{2} : 4

(6)^{2} - (2)^{2}

(6)^{2} = 6

(6)^{2}

Apply radical rule:

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

= (6^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 6^{\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

= 6

(2)^{2} = 2

(2)^{2}

Apply radical rule:

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

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\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

= 2

= 6 - 2

Subtract the numbers:

6 - 2 = 4

= 4

= 4

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

Divide the numbers:

\frac{4}{4} = 1

= 6 + 2

= 6 + 2

= 6 + 2

sec (4 0^{\circ} + 2 θ) = 6 + 2

Apply trig inverse properties

sec (4 0^{\circ} + 2 θ) = 6 + 2

General solutions for

sec (4 0^{\circ} + 2 θ) = 6 + 2

sec (x) = a \Rightarrow x = arcsec (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arcsec (a) + 36 0^{\circ} n

4 0^{\circ} + 2 θ = arcsec (6 + 2) + 36 0^{\circ} n, 4 0^{\circ} + 2 θ = 36 0^{\circ} - arcsec (6 + 2) + 36 0^{\circ} n

4 0^{\circ} + 2 θ = arcsec (6 + 2) + 36 0^{\circ} n, 4 0^{\circ} + 2 θ = 36 0^{\circ} - arcsec (6 + 2) + 36 0^{\circ} n

Solve

4 0^{\circ} + 2 θ = arcsec (6 + 2) + 36 0^{\circ} n : θ = 18 0^{\circ} n - 2 0^{\circ} + \frac{arcsec ( 6 + 2 )}{2}

4 0^{\circ} + 2 θ = arcsec (6 + 2) + 36 0^{\circ} n

Move

4 0^{\circ}

to the right side

4 0^{\circ} + 2 θ = arcsec (6 + 2) + 36 0^{\circ} n

Subtract

4 0^{\circ}

from both sides

4 0^{\circ} + 2 θ - 4 0^{\circ} = arcsec (6 + 2) + 36 0^{\circ} n - 4 0^{\circ}

Simplify

2 θ = arcsec (6 + 2) + 36 0^{\circ} n - 4 0^{\circ}

2 θ = arcsec (6 + 2) + 36 0^{\circ} n - 4 0^{\circ}

Divide both sides by

2

2 θ = arcsec (6 + 2) + 36 0^{\circ} n - 4 0^{\circ}

Divide both sides by

2

\frac{2 θ}{2} = \frac{arcsec ( 6 + 2 )}{2} + \frac{36 0 ^{\circ} n}{2} - \frac{4 0 ^{\circ}}{2}

Simplify

\frac{2 θ}{2} = \frac{arcsec ( 6 + 2 )}{2} + \frac{36 0 ^{\circ} n}{2} - \frac{4 0 ^{\circ}}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{arcsec ( 6 + 2 )}{2} + \frac{36 0 ^{\circ} n}{2} - \frac{4 0 ^{\circ}}{2} : 18 0^{\circ} n - 2 0^{\circ} + \frac{arcsec ( 6 + 2 )}{2}

\frac{arcsec ( 6 + 2 )}{2} + \frac{36 0 ^{\circ} n}{2} - \frac{4 0 ^{\circ}}{2}

Group like terms

= \frac{36 0 ^{\circ} n}{2} - \frac{4 0 ^{\circ}}{2} + \frac{arcsec ( 6 + 2 )}{2}

\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

\frac{4 0 ^{\circ}}{2} = 2 0^{\circ}

\frac{4 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{36 0 ^{\circ}}{9 \cdot 2}

Multiply the numbers:

9 \cdot 2 = 18

= 2 0^{\circ}

Cancel the common factor:

2

= 2 0^{\circ}

= 18 0^{\circ} n - 2 0^{\circ} + \frac{arcsec ( 6 + 2 )}{2}

θ = 18 0^{\circ} n - 2 0^{\circ} + \frac{arcsec ( 6 + 2 )}{2}

θ = 18 0^{\circ} n - 2 0^{\circ} + \frac{arcsec ( 6 + 2 )}{2}

θ = 18 0^{\circ} n - 2 0^{\circ} + \frac{arcsec ( 6 + 2 )}{2}

Solve

4 0^{\circ} + 2 θ = 36 0^{\circ} - arcsec (6 + 2) + 36 0^{\circ} n : θ = 18 0^{\circ} + 18 0^{\circ} n - 2 0^{\circ} - \frac{arcsec ( 6 + 2 )}{2}

4 0^{\circ} + 2 θ = 36 0^{\circ} - arcsec (6 + 2) + 36 0^{\circ} n

Move

4 0^{\circ}

to the right side

4 0^{\circ} + 2 θ = 36 0^{\circ} - arcsec (6 + 2) + 36 0^{\circ} n

Subtract

4 0^{\circ}

from both sides

4 0^{\circ} + 2 θ - 4 0^{\circ} = 36 0^{\circ} - arcsec (6 + 2) + 36 0^{\circ} n - 4 0^{\circ}

Simplify

2 θ = 36 0^{\circ} - arcsec (6 + 2) + 36 0^{\circ} n - 4 0^{\circ}

2 θ = 36 0^{\circ} - arcsec (6 + 2) + 36 0^{\circ} n - 4 0^{\circ}

Divide both sides by

2

2 θ = 36 0^{\circ} - arcsec (6 + 2) + 36 0^{\circ} n - 4 0^{\circ}

Divide both sides by

2

\frac{2 θ}{2} = 18 0^{\circ} - \frac{arcsec ( 6 + 2 )}{2} + \frac{36 0 ^{\circ} n}{2} - \frac{4 0 ^{\circ}}{2}

Simplify

\frac{2 θ}{2} = 18 0^{\circ} - \frac{arcsec ( 6 + 2 )}{2} + \frac{36 0 ^{\circ} n}{2} - \frac{4 0 ^{\circ}}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

18 0^{\circ} - \frac{arcsec ( 6 + 2 )}{2} + \frac{36 0 ^{\circ} n}{2} - \frac{4 0 ^{\circ}}{2} : 18 0^{\circ} + 18 0^{\circ} n - 2 0^{\circ} - \frac{arcsec ( 6 + 2 )}{2}

18 0^{\circ} - \frac{arcsec ( 6 + 2 )}{2} + \frac{36 0 ^{\circ} n}{2} - \frac{4 0 ^{\circ}}{2}

Group like terms

= 18 0^{\circ} + \frac{36 0 ^{\circ} n}{2} - \frac{4 0 ^{\circ}}{2} - \frac{arcsec ( 6 + 2 )}{2}

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

18 0^{\circ}

Divide the numbers:

\frac{2}{2} = 1

= 18 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

\frac{4 0 ^{\circ}}{2} = 2 0^{\circ}

\frac{4 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{36 0 ^{\circ}}{9 \cdot 2}

Multiply the numbers:

9 \cdot 2 = 18

= 2 0^{\circ}

Cancel the common factor:

2

= 2 0^{\circ}

= 18 0^{\circ} + 18 0^{\circ} n - 2 0^{\circ} - \frac{arcsec ( 6 + 2 )}{2}

θ = 18 0^{\circ} + 18 0^{\circ} n - 2 0^{\circ} - \frac{arcsec ( 6 + 2 )}{2}

θ = 18 0^{\circ} + 18 0^{\circ} n - 2 0^{\circ} - \frac{arcsec ( 6 + 2 )}{2}

θ = 18 0^{\circ} + 18 0^{\circ} n - 2 0^{\circ} - \frac{arcsec ( 6 + 2 )}{2}

θ = 18 0^{\circ} n - 2 0^{\circ} + \frac{arcsec ( 6 + 2 )}{2}, θ = 18 0^{\circ} + 18 0^{\circ} n - 2 0^{\circ} - \frac{arcsec ( 6 + 2 )}{2}

Show solutions in decimal form

θ = 18 0^{\circ} n - 2 0^{\circ} + \frac{1.30899 \dots}{2}, θ = 18 0^{\circ} + 18 0^{\circ} n - 2 0^{\circ} - \frac{1.30899 \dots}{2}

Popular Examples

arcsin(x)-arccos(x)=arcsin(1/2)

arcsin (x) - arccos (x) = arcsin (\frac{1}{2})

cos^2(x)-cos(x)=sin^2(x)

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

0=sec^2(x)

0 = sec^{2} (x)

2+2cos(θ)=6cos(θ)

2 + 2 cos (θ) = 6 cos (θ)

3sin^2(x)+sin(x)-4=0

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

Frequently Asked Questions (FAQ)

What is the general solution for sec(40+2θ)=csc(15) ?
The general solution for sec(40+2θ)=csc(15) is θ=180n-20+(1.30899…)/2 ,θ=180+180n-20-(1.30899…)/2