1/(1-sin(x))+1/(1+sin(x))=2csc^2(x)

Solution

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

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solution steps

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

Subtract

2 csc^{2} (x)

from both sides

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

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

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

Rewrite using trig identities

4 - 2 csc^{2} (x) = 0

Solve by substitution

csc (x) = 2, csc (x) = - 2

csc (x) = 2 : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

csc (x) = - 2 : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

cos (θ) = - \frac{2}{5}, tan (2 θ)

sin (x + \frac{π}{6}) = \frac{2}{2}

\frac{1}{2} = cos (x)

2 sin (4 x) = sin (2 x)