What is the general solution for 1/(tan^2(x))-cos^2(x)=cos^2(x)-cot^2(x) ?

The general solution for 1/(tan^2(x))-cos^2(x)=cos^2(x)-cot^2(x) is No Solution for x\in\mathbb{R}

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1/(tan^2(x))-cos^2(x)=cos^2(x)-cot^2(x)

\frac{1}{tan ^{2} ( x )} - cos^{2} (x) = cos^{2} (x) - cot^{2} (x)

No Solution for x \in R

Solution steps

\frac{1}{tan ^{2} ( x )} - cos^{2} (x) = cos^{2} (x) - cot^{2} (x)

Subtract

cos^{2} (x) - cot^{2} (x)

from both sides

\frac{1}{tan ^{2} ( x )} - 2 cos^{2} (x) + cot^{2} (x) = 0

Simplify

\frac{1}{tan ^{2} ( x )} - 2 cos^{2} (x) + cot^{2} (x) : \frac{1 - 2 cos ^{2} ( x ) tan ^{2} ( x ) + cot ^{2} ( x ) tan ^{2} ( x )}{tan ^{2} ( x )}

\frac{1 - 2 cos ^{2} ( x ) tan ^{2} ( x ) + cot ^{2} ( x ) tan ^{2} ( x )}{tan ^{2} ( x )} = 0

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

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

Express with sin, cos

2 - 2 sin^{2} (x) = 0

Solve by substitution

sin (x) = 1, sin (x) = - 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

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

No Solution for x \in R

3 ∣ sin (3 θ) ∣ = 0

- 2 cos (x) - 4 sin (2 x) = 0

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

33 cos (x) - 6 sin (x) cos (x) = 0

2 tan^{2} (x) + 4 sec (x) + sec^{2} (x) = 2

What is the general solution for 1/(tan^2(x))-cos^2(x)=cos^2(x)-cot^2(x) ?
The general solution for 1/(tan^2(x))-cos^2(x)=cos^2(x)-cot^2(x) is No Solution for x\in\mathbb{R}