((1-cos(x))(csc^2(x)-4))/(1+cos(x))=0

Solution

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

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solution steps

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

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

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

Solving each part separately

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

1 - cos (x) = 0 : x = 2 πn

csc^{2} (x) - 4 = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Since the equation is undefined for:

2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

6 sin (a) + 4 = 2 sin (a) + 8

sin (3 x) = 0, - 2 < x < 2

tan (x) = \frac{2}{6}

2 sin^{2} (x) - 5 sin (x) = 3

cos (8 x) - cos (2 x) = sin (5 x)