What is the general solution for sec^2(θ)cot(θ)=8 ?

The general solution for sec^2(θ)cot(θ)=8 is θ=(0.25268…)/2+pin,θ= pi/2-(0.25268…)/2+pin

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sec^2(θ)cot(θ)=8

sec^{2} (θ) cot (θ) = 8

θ = \frac{0.25268 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.25268 \dots}{2} + πn

Solution steps

sec^{2} (θ) cot (θ) = 8

Subtract

8

from both sides

sec^{2} (θ) cot (θ) - 8 = 0

Express with sin, cos

\frac{1 - 8 cos ( θ ) sin ( θ )}{cos ( θ ) sin ( θ )} = 0

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

1 - 8 cos (θ) sin (θ) = 0

Rewrite using trig identities

1 - 8 \cdot \frac{sin ( 2 θ )}{2} = 0

8 \cdot \frac{sin ( 2 θ )}{2} = 4 sin (2 θ)

1 - 4 sin (2 θ) = 0

Move

1

to the right side

- 4 sin (2 θ) = - 1

Divide both sides by

- 4

sin (2 θ) = \frac{1}{4}

Apply trig inverse properties

2 θ = arcsin (\frac{1}{4}) + 2 πn, 2 θ = π - arcsin (\frac{1}{4}) + 2 πn

Solve

2 θ = arcsin (\frac{1}{4}) + 2 πn : θ = \frac{arcsin ( \frac{1}{4} )}{2} + πn

Solve

2 θ = π - arcsin (\frac{1}{4}) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

θ = \frac{arcsin ( \frac{1}{4} )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.25268 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.25268 \dots}{2} + πn

4 + 5 sin (x) = 1

120 = 23 sin (\frac{π}{6} (x)) + 107

4 sin (3 θ) = 2

cos (x) = cos (3 x) + 2 sin (2 x)

csc (x) (2 sin (x) - 2) = 0

What is the general solution for sec^2(θ)cot(θ)=8 ?
The general solution for sec^2(θ)cot(θ)=8 is θ=(0.25268…)/2+pin,θ= pi/2-(0.25268…)/2+pin