What is the general solution for sec^4(x)=sec^2(x)tan^2(x)-2tan^4(x) ?

The general solution for sec^4(x)=sec^2(x)tan^2(x)-2tan^4(x) is No Solution for x\in\mathbb{R}

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sec^4(x)=sec^2(x)tan^2(x)-2tan^4(x)

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

No Solution for x \in R

Solution steps

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

Subtract

sec^{2} (x) tan^{2} (x) - 2 tan^{4} (x)

from both sides

sec^{4} (x) - sec^{2} (x) tan^{2} (x) + 2 tan^{4} (x) = 0

Express with sin, cos

\frac{1 - sin ^{2} ( x ) + 2 sin ^{4} ( x )}{cos ^{4} ( x )} = 0

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

1 - sin^{2} (x) + 2 sin^{4} (x) = 0

Solve by substitution

sin (x) = \frac{7}{2 2 2 2 - 1} + \frac{2 2 - 1}{2 2} i, sin (x) = - \frac{7}{2 2 2 2 - 1} - \frac{2 2 - 1}{2 2} i, sin (x) = - \frac{7}{2 2 2 2 - 1} + \frac{2 2 - 1}{2 2} i, sin (x) = \frac{7}{2 2 2 2 - 1} - \frac{2 2 - 1}{2 2} i

sin (x) = \frac{7}{2 2 2 2 - 1} + \frac{2 2 - 1}{2 2} i :

No Solution

sin (x) = - \frac{7}{2 2 2 2 - 1} - \frac{2 2 - 1}{2 2} i :

No Solution

sin (x) = - \frac{7}{2 2 2 2 - 1} + \frac{2 2 - 1}{2 2} i :

No Solution

sin (x) = \frac{7}{2 2 2 2 - 1} - \frac{2 2 - 1}{2 2} i :

No Solution

Combine all the solutions

No Solution for x \in R

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0.08 = 0.1 cos (4 x - 1.57)

cos (2 t) - sin (t) = 0.5, 0 < t < 2 π

25 sin (2 x) - 50 cos (x) = 0

What is the general solution for sec^4(x)=sec^2(x)tan^2(x)-2tan^4(x) ?
The general solution for sec^4(x)=sec^2(x)tan^2(x)-2tan^4(x) is No Solution for x\in\mathbb{R}