What is the general solution for cos^2(β)-cot^2(β)=cos^2(β)cot^2(β) ?

The general solution for cos^2(β)-cot^2(β)=cos^2(β)cot^2(β) is β= pi/2+pin

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cos^2(β)-cot^2(β)=cos^2(β)cot^2(β)

cos^{2} (β) - cot^{2} (β) = cos^{2} (β) cot^{2} (β)

β = \frac{π}{2} + πn

Solution steps

cos^{2} (β) - cot^{2} (β) = cos^{2} (β) cot^{2} (β)

Subtract

cos^{2} (β) cot^{2} (β)

from both sides

cos^{2} (β) - cot^{2} (β) - cos^{2} (β) cot^{2} (β) = 0

Express with sin, cos

\frac{- cos ^{2} ( β ) - cos ^{4} ( β ) + cos ^{2} ( β ) sin ^{2} ( β )}{sin ^{2} ( β )} = 0

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

- cos^{2} (β) - cos^{4} (β) + cos^{2} (β) sin^{2} (β) = 0

Factor

- cos^{2} (β) - cos^{4} (β) + cos^{2} (β) sin^{2} (β) : - cos^{2} (β) (1 + cos^{2} (β) - sin^{2} (β))

- cos^{2} (β) (1 + cos^{2} (β) - sin^{2} (β)) = 0

Solving each part separately

cos^{2} (β) = 0 or 1 + cos^{2} (β) - sin^{2} (β) = 0

cos^{2} (β) = 0 : β = \frac{π}{2} + 2 πn, β = \frac{3 π}{2} + 2 πn

1 + cos^{2} (β) - sin^{2} (β) = 0 : β = \frac{π}{2} + πn

Combine all the solutions

β = \frac{π}{2} + 2 πn, β = \frac{3 π}{2} + 2 πn, β = \frac{π}{2} + πn

Merge Overlapping Intervals

β = \frac{π}{2} + πn

4 tan (x) + 4 = 0, 0 \leq x < 2 π

cos^{3} (x) = cos (x), 0 \leq x < 2 π

3 - 2 sin^{2} (x) = 2 sin^{2} (\frac{x}{2})

cot (θ) = - \frac{7}{5}, \frac{3 π}{2} < θ < 2 π, tan (θ)

sin (r) = 0.58

What is the general solution for cos^2(β)-cot^2(β)=cos^2(β)cot^2(β) ?
The general solution for cos^2(β)-cot^2(β)=cos^2(β)cot^2(β) is β= pi/2+pin