What is the general solution for sec^2(x)cot(x)=8,0<x<pi ?

The general solution for sec^2(x)cot(x)=8,0<x<pi is x=(0.25268…)/2 ,x=(pi-0.25268…)/2

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sec^2(x)cot(x)=8,0<x<pi

sec^{2} (x) cot (x) = 8, 0 < x < π

x = \frac{0.25268 \dots}{2}, x = \frac{π - 0.25268 \dots}{2}

Solution steps

sec^{2} (x) cot (x) = 8, 0 < x < π

Subtract

8

from both sides

sec^{2} (x) cot (x) - 8 = 0

Express with sin, cos

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

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

1 - 8 cos (x) sin (x) = 0

Rewrite using trig identities

1 - 8 \cdot \frac{sin ( 2 x )}{2} = 0

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

1 - 4 sin (2 x) = 0

Move

1

to the right side

- 4 sin (2 x) = - 1

Divide both sides by

- 4

sin (2 x) = \frac{1}{4}

Apply trig inverse properties

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

Solve

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

Solve

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

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

Solutions for the range

0 < x < π

x = \frac{arcsin ( \frac{1}{4} )}{2}, x = \frac{π - arcsin ( \frac{1}{4} )}{2}

Show solutions in decimal form

x = \frac{0.25268 \dots}{2}, x = \frac{π - 0.25268 \dots}{2}

tan (x) = \frac{1.5}{2}

3 csc^{2} (x) - 5 csc (x) = 2

tan (2 x) - 3 tan (x) = 0

4 sin (x) - 3 cos (x) = 0

25 cos^{2} (3 x) + 10 cos (3 x) = 15

What is the general solution for sec^2(x)cot(x)=8,0<x<pi ?
The general solution for sec^2(x)cot(x)=8,0<x<pi is x=(0.25268…)/2 ,x=(pi-0.25268…)/2