What is the general solution for 8tan(θ/2)+8cos(θ)tan(θ/2)=1 ?

The general solution for 8tan(θ/2)+8cos(θ)tan(θ/2)=1 is θ=0.12532…+2pin,θ=pi-0.12532…+2pin

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8tan(θ/2)+8cos(θ)tan(θ/2)=1

Solution

8 tan (\frac{θ}{2}) + 8 cos (θ) tan (\frac{θ}{2}) = 1

Solution

θ = 0.12532 \dots + 2 πn, θ = π - 0.12532 \dots + 2 πn

Degrees

θ = 7.18075 \dots^{\circ} + 36 0^{\circ} n, θ = 172.81924 \dots^{\circ} + 36 0^{\circ} n

Solution steps

8 tan (\frac{θ}{2}) + 8 cos (θ) tan (\frac{θ}{2}) = 1

Subtract

1

from both sides

8 tan (\frac{θ}{2}) + 8 cos (θ) tan (\frac{θ}{2}) - 1 = 0

Let:

u = \frac{θ}{2}

8 tan (u) + 8 cos (2 u) tan (u) - 1 = 0

Rewrite using trig identities

- 1 + 8 tan (u) + 8 cos (2 u) tan (u)

Use the Double Angle identity:

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

= - 1 + 8 tan (u) + 8 (2 cos^{2} (u) - 1) tan (u)

Simplify

- 1 + 8 tan (u) + 8 (2 cos^{2} (u) - 1) tan (u) : 16 cos^{2} (u) tan (u) - 1

- 1 + 8 tan (u) + 8 (2 cos^{2} (u) - 1) tan (u)

= - 1 + 8 tan (u) + 8 tan (u) (2 cos^{2} (u) - 1)

Expand

8 tan (u) (2 cos^{2} (u) - 1) : 16 cos^{2} (u) tan (u) - 8 tan (u)

8 tan (u) (2 cos^{2} (u) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 8 tan (u), b = 2 cos^{2} (u), c = 1

= 8 tan (u) \cdot 2 cos^{2} (u) - 8 tan (u) \cdot 1

= 8 \cdot 2 cos^{2} (u) tan (u) - 8 \cdot 1 \cdot tan (u)

Simplify

8 \cdot 2 cos^{2} (u) tan (u) - 8 \cdot 1 \cdot tan (u) : 16 cos^{2} (u) tan (u) - 8 tan (u)

8 \cdot 2 cos^{2} (u) tan (u) - 8 \cdot 1 \cdot tan (u)

Multiply the numbers:

8 \cdot 2 = 16

= 16 cos^{2} (u) tan (u) - 8 \cdot 1 \cdot tan (u)

Multiply the numbers:

8 \cdot 1 = 8

= 16 cos^{2} (u) tan (u) - 8 tan (u)

= 16 cos^{2} (u) tan (u) - 8 tan (u)

= - 1 + 8 tan (u) + 16 cos^{2} (u) tan (u) - 8 tan (u)

Simplify

- 1 + 8 tan (u) + 16 cos^{2} (u) tan (u) - 8 tan (u) : 16 cos^{2} (u) tan (u) - 1

- 1 + 8 tan (u) + 16 cos^{2} (u) tan (u) - 8 tan (u)

Group like terms

= 8 tan (u) + 16 cos^{2} (u) tan (u) - 8 tan (u) - 1

Add similar elements:

8 tan (u) - 8 tan (u) = 0

= 16 cos^{2} (u) tan (u) - 1

= 16 cos^{2} (u) tan (u) - 1

= 16 cos^{2} (u) tan (u) - 1

Use the basic trigonometric identity:

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

= - 1 + 16 cos^{2} (u) \frac{sin ( u )}{cos ( u )}

16 cos^{2} (u) \frac{sin ( u )}{cos ( u )} = 16 sin (u) cos (u)

16 cos^{2} (u) \frac{sin ( u )}{cos ( u )}

Multiply fractions:

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

= \frac{sin ( u ) \cdot 16 cos ^{2} ( u )}{cos ( u )}

Cancel the common factor:

cos (u)

= 16 sin (u) cos (u)

= - 1 + 16 sin (u) cos (u)

Use the Double Angle identity:

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

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

= - 1 + 16 \cdot \frac{sin ( 2 u )}{2}

- 1 + 16 \cdot \frac{sin ( 2 u )}{2} = 0

16 \cdot \frac{sin ( 2 u )}{2} = 8 sin (2 u)

16 \cdot \frac{sin ( 2 u )}{2}

Multiply fractions:

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

= \frac{sin ( 2 u ) \cdot 16}{2}

Divide the numbers:

\frac{16}{2} = 8

= 8 sin (2 u)

- 1 + 8 sin (2 u) = 0

Move

1

to the right side

- 1 + 8 sin (2 u) = 0

Add

1

to both sides

- 1 + 8 sin (2 u) + 1 = 0 + 1

Simplify

8 sin (2 u) = 1

8 sin (2 u) = 1

Divide both sides by

8

8 sin (2 u) = 1

Divide both sides by

8

\frac{8 sin ( 2 u )}{8} = \frac{1}{8}

Simplify

sin (2 u) = \frac{1}{8}

sin (2 u) = \frac{1}{8}

Apply trig inverse properties

sin (2 u) = \frac{1}{8}

General solutions for

sin (2 u) = \frac{1}{8}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

2 u = arcsin (\frac{1}{8}) + 2 πn, 2 u = π - arcsin (\frac{1}{8}) + 2 πn

2 u = arcsin (\frac{1}{8}) + 2 πn, 2 u = π - arcsin (\frac{1}{8}) + 2 πn

Solve

2 u = arcsin (\frac{1}{8}) + 2 πn : u = \frac{arcsin ( \frac{1}{8} )}{2} + πn

2 u = arcsin (\frac{1}{8}) + 2 πn

Divide both sides by

2

2 u = arcsin (\frac{1}{8}) + 2 πn

Divide both sides by

2

\frac{2 u}{2} = \frac{arcsin ( \frac{1}{8} )}{2} + \frac{2 πn}{2}

Simplify

u = \frac{arcsin ( \frac{1}{8} )}{2} + πn

u = \frac{arcsin ( \frac{1}{8} )}{2} + πn

Solve

2 u = π - arcsin (\frac{1}{8}) + 2 πn : u = \frac{π}{2} - \frac{arcsin ( \frac{1}{8} )}{2} + πn

2 u = π - arcsin (\frac{1}{8}) + 2 πn

Divide both sides by

2

2 u = π - arcsin (\frac{1}{8}) + 2 πn

Divide both sides by

2

\frac{2 u}{2} = \frac{π}{2} - \frac{arcsin ( \frac{1}{8} )}{2} + \frac{2 πn}{2}

Simplify

u = \frac{π}{2} - \frac{arcsin ( \frac{1}{8} )}{2} + πn

u = \frac{π}{2} - \frac{arcsin ( \frac{1}{8} )}{2} + πn

u = \frac{arcsin ( \frac{1}{8} )}{2} + πn, u = \frac{π}{2} - \frac{arcsin ( \frac{1}{8} )}{2} + πn

Substitute back

u = \frac{θ}{2}

\frac{θ}{2} = \frac{arcsin ( \frac{1}{8} )}{2} + πn : θ = arcsin (\frac{1}{8}) + 2 πn

\frac{θ}{2} = \frac{arcsin ( \frac{1}{8} )}{2} + πn

Multiply both sides by

2

\frac{θ}{2} = \frac{arcsin ( \frac{1}{8} )}{2} + πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{arcsin ( \frac{1}{8} )}{2} + 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{arcsin ( \frac{1}{8} )}{2} + 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{arcsin ( \frac{1}{8} )}{2} + 2 πn : arcsin (\frac{1}{8}) + 2 πn

2 \cdot \frac{arcsin ( \frac{1}{8} )}{2} + 2 πn

2 \cdot \frac{arcsin ( \frac{1}{8} )}{2} = arcsin (\frac{1}{8})

2 \cdot \frac{arcsin ( \frac{1}{8} )}{2}

Multiply fractions:

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

= \frac{arcsin ( \frac{1}{8} ) \cdot 2}{2}

Cancel the common factor:

2

= arcsin (\frac{1}{8})

= arcsin (\frac{1}{8}) + 2 πn

θ = arcsin (\frac{1}{8}) + 2 πn

θ = arcsin (\frac{1}{8}) + 2 πn

θ = arcsin (\frac{1}{8}) + 2 πn

\frac{θ}{2} = \frac{π}{2} - \frac{arcsin ( \frac{1}{8} )}{2} + πn : θ = π - arcsin (\frac{1}{8}) + 2 πn

\frac{θ}{2} = \frac{π}{2} - \frac{arcsin ( \frac{1}{8} )}{2} + πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{2} - \frac{arcsin ( \frac{1}{8} )}{2} + πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{2} - 2 \cdot \frac{arcsin ( \frac{1}{8} )}{2} + 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{π}{2} - 2 \cdot \frac{arcsin ( \frac{1}{8} )}{2} + 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{π}{2} - 2 \cdot \frac{arcsin ( \frac{1}{8} )}{2} + 2 πn : π - arcsin (\frac{1}{8}) + 2 πn

2 \cdot \frac{π}{2} - 2 \cdot \frac{arcsin ( \frac{1}{8} )}{2} + 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

Multiply fractions:

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

= \frac{π 2}{2}

Cancel the common factor:

2

= π

2 \cdot \frac{arcsin ( \frac{1}{8} )}{2} = arcsin (\frac{1}{8})

2 \cdot \frac{arcsin ( \frac{1}{8} )}{2}

Multiply fractions:

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

= \frac{arcsin ( \frac{1}{8} ) \cdot 2}{2}

Cancel the common factor:

2

= arcsin (\frac{1}{8})

= π - arcsin (\frac{1}{8}) + 2 πn

θ = π - arcsin (\frac{1}{8}) + 2 πn

θ = π - arcsin (\frac{1}{8}) + 2 πn

θ = π - arcsin (\frac{1}{8}) + 2 πn

θ = arcsin (\frac{1}{8}) + 2 πn, θ = π - arcsin (\frac{1}{8}) + 2 πn

Show solutions in decimal form

θ = 0.12532 \dots + 2 πn, θ = π - 0.12532 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 8tan(θ/2)+8cos(θ)tan(θ/2)=1 ?
The general solution for 8tan(θ/2)+8cos(θ)tan(θ/2)=1 is θ=0.12532…+2pin,θ=pi-0.12532…+2pin