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

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

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

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

No Solution for x \in R

Solution steps

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

Subtract

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

from both sides

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

Simplify

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

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

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

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

Express with sin, cos

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

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

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

Rewrite using trig identities

- 2 + sin (x) + 1 = 0

Group like terms

sin (x) - 2 + 1 = 0

Add/Subtract the numbers:

- 2 + 1 = - 1

sin (x) - 1 = 0

Move

1

to the right side

sin (x) = 1

General solutions for

sin (x) = 1

x = \frac{π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn

No Solution for x \in R

42 tan (x) - 2 = 32 tan (x)

7 sin (2 x) + 12 cos (x) = 0

θ, cos (θ) = - cos (4 0^{\circ})

- 1 - 2 sec^{2} (x) = - 3 sec^{2} (x)

tanh (x + 1) + 1 = 2

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