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

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

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(1-sin(x))(((1+sin(x)))/(2cos(x)))=1

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

No Solution for x \in R

Solution steps

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

Subtract

1

from both sides

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

Simplify

\frac{( 1 + sin ( x ) ) ( 1 - sin ( x ) )}{2 cos ( x )} - 1 : \frac{( 1 + sin ( x ) ) ( 1 - sin ( x ) ) - 2 cos ( x )}{2 cos ( x )}

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

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

(1 + sin (x)) (1 - sin (x)) - 2 cos (x) = 0

Rewrite using trig identities

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

Solve by substitution

cos (x) = 2, cos (x) = 0

cos (x) = 2 :

No Solution

cos (x) = 0 : x = \frac{π}{2} + 2 πn, 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

\frac{9.03}{sin ( 4 0 ^{\circ} )} = \frac{10}{sin ( x )}

tan (θ) (sec (θ) - 1) = 0

tan (x) + sec (x) = 2

sin (x) = \frac{4}{5}, \frac{π}{2} \leq x \leq π

2 - 2 cosh (t - 2) = 0

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