What is the general solution for 1+cot^2(x)=8sin(x) ?

The general solution for 1+cot^2(x)=8sin(x) is x= pi/6+2pin,x=(5pi)/6+2pin

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1+cot^2(x)=8sin(x)

1 + cot^{2} (x) = 8 sin (x)

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solution steps

1 + cot^{2} (x) = 8 sin (x)

Subtract

8 sin (x)

from both sides

1 + cot^{2} (x) - 8 sin (x) = 0

Rewrite using trig identities

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

Solve by substitution

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{4} + i \frac{3}{4}, sin (x) = - \frac{1}{4} - i \frac{3}{4}

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = - \frac{1}{4} + i \frac{3}{4} :

No Solution

sin (x) = - \frac{1}{4} - i \frac{3}{4} :

No Solution

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

\frac{cos ( x ) - 3 sin ( x )}{sin ( x ) + 3 cos ( x )} = - \frac{1}{2}

cos (\frac{π}{2} - x) = 0

1 sin (4 5^{\circ}) = 2.42 sin (r)

sec (x) = 4 sin (x)

What is the general solution for 1+cot^2(x)=8sin(x) ?
The general solution for 1+cot^2(x)=8sin(x) is x= pi/6+2pin,x=(5pi)/6+2pin