What is the general solution for (h(sin^2(x)))/((cos(x)-1))=0 ?

The general solution for (h(sin^2(x)))/((cos(x)-1))=0 is x=pi+2pin

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(h(sin^2(x)))/((cos(x)-1))=0

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

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

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

h sin^{2} (x) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

sin^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Since the equation is undefined for:

2 πn

x = π + 2 πn

cos (a) = \frac{- 3}{5}

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

sin^{22} (x) = \frac{1}{2}

cos^{5} (x) = cos (x)

tan (x) = (\frac{1.8}{3.6})

What is the general solution for (h(sin^2(x)))/((cos(x)-1))=0 ?
The general solution for (h(sin^2(x)))/((cos(x)-1))=0 is x=pi+2pin