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

The general solution for (1(cos^2(x)))/((1-sin^2(x)))=0 is No Solution for x\in\mathbb{R}

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

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

No Solution for x \in R

Solution steps

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

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

1 \cdot cos^{2} (x) = 0

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

cos^{2} (x) = 0

Apply rule

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

cos (x) = 0

General solutions for

cos (x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

sin (x) sin^{3} (x) - sin^{5} (x) sin^{3} (x) = 0

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

3 cos^{2} (x) + 5 sin (x) - 4 = 0

x, d^{2} + 3 d + 2 y = 4 cos^{2} (x)

(1 + tan^{2} (x)) cos^{2} (x) = sec (x)

What is the general solution for (1(cos^2(x)))/((1-sin^2(x)))=0 ?
The general solution for (1(cos^2(x)))/((1-sin^2(x)))=0 is No Solution for x\in\mathbb{R}