What is the general solution for (tan^2(x))/(sec(x)+1)=tan(x) ?

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

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(tan^2(x))/(sec(x)+1)=tan(x)

\frac{tan ^{2} ( x )}{sec ( x ) + 1} = tan (x)

No Solution for x \in R

Solution steps

\frac{tan ^{2} ( x )}{sec ( x ) + 1} = tan (x)

Subtract

tan (x)

from both sides

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

Simplify

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

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

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

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

Factor

tan^{2} (x) - tan (x) (sec (x) + 1) : tan (x) (tan (x) - 1 - sec (x))

tan (x) (tan (x) - 1 - sec (x)) = 0

Solving each part separately

tan (x) = 0 or tan (x) - 1 - sec (x) = 0

tan (x) = 0 : x = πn

tan (x) - 1 - sec (x) = 0 : x = 2 πn + π

Combine all the solutions

x = πn, x = 2 πn + π

Since the equation is undefined for:

πn, 2 πn + π

No Solution for x \in R

2 cos^{2} (x) + cos (x) = 1, 0 \leq x < 2 π

sin (β) = - 0, 8 (θ \in β v c) s = sec (β) - tan (β)

cos (\frac{2 x - π}{17}) = 0

tan (X) cot (X) - tan (X) + 2 cot (X) = 0

sin (x) = - 0.3926

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