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

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

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

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

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

- 1 + 8 sin (2 u) = 0

Move

1

to the right side

8 sin (2 u) = 1

Divide both sides by

8

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

Apply trig inverse properties

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

Solve

2 u = π - 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{π}{2} - \frac{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

m (0) = sin (2 x) + 2

6 cos^{2} (x) + cos (x) - 1 = 0

sin^{2} (x) + 2 cos (x) + 2 = 0

6 sin^{2} (x) - 5 cos (x) - 2 = 0

sin (θ) + cos (θ) = \frac{3}{5}

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